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what is a cell right now your body is doing a million things at once . it ’ s sending electrical impulses , pumping blood , filtering urine , digesting food , making protein , storing fat , and that ’ s just the stuff you ’ re not thinking about ! you can do all this because you are made of cells — tiny units of life that are like specialized factories , full of machinery designed to accomplish the business of life . cells make up every living thing , from blue whales to the archaebacteria that live inside volcanos . just like the organisms they make up , cells can come in all shapes and sizes . nerve cells in giant squids can reach up to 12m [ 39 ft ] in length , while human eggs ( the largest human cells ) are about 0.1mm across . plant cells have protective walls made of cellulose ( which also makes up the strings in celery that make it so hard to eat ) while fungal cell walls are made from the same stuff as lobster shells . however , despite this vast range in size , shape , and function , all these little factories have the same basic machinery . there are two main types of cells , prokaryotic and eukaryotic . prokaryotes are cells that do not have membrane bound nuclei , whereas eukaryotes do . the rest of our discussion will strictly be on eukaryotes . think about what a factory needs in order to function effectively . at its most basic , a factory needs a building , a product , and a way to make that product . all cells have membranes ( the building ) , dna ( the various blueprints ) , and ribosomes ( the production line ) , and so are able to make proteins ( the product - let ’ s say we ’ re making toys ) . this article will focus on eukaryotes , since they are the cell type that contains organelles . what ’ s found inside a cell an organelle ( think of it as a cell ’ s internal organ ) is a membrane bound structure found within a cell . just like cells have membranes to hold everything in , these mini-organs are also bound in a double layer of phospholipids to insulate their little compartments within the larger cells . you can think of organelles as smaller rooms within the factory , with specialized conditions to help these rooms carry out their specific task ( like a break room stocked with goodies or a research room with cool gadgets and a special air filter ) . these organelles are found in the cytoplasm , a viscous liquid found within the cell membrane that houses the organelles and is the location of most of the action happening in a cell . below is a table of the organelles found in the basic human cell , which we ’ ll be using as our template for this discussion . organelle | function | factory part : - : | : - : | : - : nucleus | dna storage | room where the blueprints are kept mitochondrion | energy production | powerplant smooth endoplasmic reticulum ( ser ) | lipid production ; detoxification | accessory production - makes decorations for the toy , etc . rough endoplasmic reticulum ( rer ) | protein production ; in particular for export out of the cell | primary production line - makes the toys golgi apparatus | protein modification and export | shipping department peroxisome | lipid destruction ; contains oxidative enzymes | security and waste removal lysosome | protein destruction | recycling and security nucleus our dna has the blueprints for every protein in our body , all packaged into a neat double helix . the processes to transform dna into proteins are known as transcription and translation , and happen in different compartments within the cell . the first step , transcription , happens in the nucleus , which holds our dna . a membrane called the nuclear envelope surrounds the nucleus , and its job is to create a room within the cell to both protect the genetic information and to house all the molecules that are involved in processing and protecting that info . this membrane is actually a set of two lipid bilayers , so there are four sheets of lipids separating the inside of the nucleus from the cytoplasm . the space between the two bilayers is known as the perinuclear space . though part of the function of the nucleus is to separate the dna from the rest of the cell , molecules must still be able to move in and out ( e.g. , rna ) . proteins channels known as nuclear pores form holes in the nuclear envelope . the nucleus itself is filled with liquid ( called nucleoplasm ) and is similar in structure and function to cytoplasm . it is here within the nucleoplasm where chromosomes ( tightly packed strands of dna containing all our blueprints ) are found . a nucleus has interesting implications for how a cell responds to its environment . thanks to the added protection of the nuclear envelope , the dna is a little bit more secure from enzymes , pathogens , and potentially harmful products of fat and protein metabolism . since this is the only permanent copy of the instructions the cell has , it is very important to keep the dna in good condition . if the dna was not sequestered away , it would be vulnerable to damage by the aforementioned dangers , which would then lead to defective protein production . imagine a giant hole or coffee stain in the blueprint for your toy - all of a sudden you don ’ t have either enough or the right information to make a critical piece of the toy . the nuclear envelope also keeps molecules responsible for dna transcription and repair close to the dna itself - otherwise those molecules would diffuse across the entire cell and it would take a lot more work and luck to get anything done ! while transcription ( making a complementary strand of rna from dna ) is completed within the nucleus , translation ( making protein from rna instructions ) takes place in the cytoplasm . if there was no barrier between the transcription and translation machineries , it ’ s possible that poorly-made or unfinished rna would get turned into poorly made and potentially dangerous proteins . before an rna can exit the nucleus to be translated , it must get special modifications , in the form of a cap and tail at either end of the molecule , that act as a stamp of approval to let the cell know this piece of rna is complete and properly made . nucleolus within the nucleus is a small subspace known as the nucleolus . it is not bound by a membrane , so it is not an organelle . this space forms near the part of dna with instructions for making ribosomes , the molecules responsible for making proteins . ribosomes are assembled in the nucleolus , and exit the nucleus with nuclear pores . in our analogy , the robots making our product are made in a special corner of the blueprint room , before being released to the factory . endoplasmic reticulum endoplasmic means inside ( endo ) the cytoplasm ( plasm ) . reticulum comes from the latin word for net . basically , an endoplasmic reticulum is a plasma membrane found inside the cell that folds in on itself to create an internal space known as the lumen . this lumen is actually continuous with the perinuclear space , so we know the endoplasmic reticulum is attached to the nuclear envelope . there are actually two different endoplasmic reticuli in a cell : the smooth endoplasmic reticulum and the rough endoplasmic reticulum . the rough endoplasmic reticulum is the site of protein production ( where we make our major product - the toy ) while the smooth endoplasmic reticulum is where lipids ( fats ) are made ( accessories for the toy , but not the central product of the factory ) . rough endoplasmic reticulum the rough endoplasmic reticulum is so-called because its surface is studded with ribosomes , the molecules in charge of protein production . when a ribosome finds a specific rna segment , that segment may tell the ribosome to travel to the rough endoplasmic reticulum and embed itself . the protein created from this segment will find itself inside the lumen of the rough endoplasmic reticulum , where it folds and is tagged with a ( usually carbohydrate ) molecule in a process known as glycosylation that marks the protein for transport to the golgi apparatus . the rough endoplasmic reticulum is continuous with the nuclear envelope , and looks like a series of canals near the nucleus . proteins made in the rough endoplasmic reticulum as destined to either be a part of a membrane , or to be secreted from the cell membrane out of the cell . without an rough endoplasmic reticulum , it would be a lot harder to distinguish between proteins that should leave the cell , and proteins that should remain . thus , the rough endoplasmic reticulum helps cells specialize and allows for greater complexity in the organism . smooth endoplasmic reticulum the smooth endoplasmic reticulum makes lipids and steroids , instead of being involved in protein synthesis . these are fat-based molecules that are important in energy storage , membrane structure , and communication ( steroids can act as hormones ) . the smooth endoplasmic reticulum is also responsible for detoxifying the cell . it is more tubular than the rough endoplasmic reticulum , and is not necessarily continuous with the nuclear envelope . every cell has a smooth endoplasmic reticulum , but the amount will vary with cell function . for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum . golgi apparatus ( aka golgi body aka golgi ) we mentioned the golgi apparatus earlier when we discussed the production of proteins in the rough endoplasmic reticulum . if the smooth and rough endoplasmic reticula are how we make our product , the golgi is the mailroom that sends our product to customers . it is responsible for packing proteins from the rough endoplasmic reticulum into membrane-bound vesicles ( tiny compartments of lipid bilayer that store molecules ) which then translocate to the cell membrane . at the cell membrane , the vesicles can fuse with the larger lipid bilayer , causing the vesicle contents to either become part of the cell membrane or be released to the outside . different molecules actually have different fates upon entering the golgi . this determination is done by tagging the proteins with special sugar molecules that act as a shipping label for the protein . the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it . molecules in this pathway are often protein channels which allow molecules into or out of the cell , or cell identifiers which project into the extracellular space and act like a name tag for the cell . secretion : some proteins are meant to be secreted from the cell to act on other parts of the body . before these vesicles can fuse with the cell membrane , they must accumulate in number , and require a special chemical signal to be released . this way shipments only go out if they ’ re worth the cost of sending them ( you generally wouldn ’ t ship just one toy and expect to profit ) . lysosome : the final destination for proteins coming through the golgi is the lysosome . vesicles sent to this acidic organelle contain enzymes that will hydrolyze the lysosome ’ s content . lysosome the lysosome is the cell ’ s recycling center . these organelles are spheres full of enzymes ready to hydrolyze ( chop up the chemical bonds of ) whatever substance crosses the membrane , so the cell can reuse the raw material . these disposal enzymes only function properly in environments with a ph of 5 , two orders of magnitude more acidic than the cell ’ s internal ph of 7 . lysosomal proteins only being active in an acidic environment acts as safety mechanism for the rest of the cell - if the lysosome were to somehow leak or burst , the degradative enzymes would inactivate before they chopped up proteins the cell still needed . peroxisome like the lysosome , the peroxisome is a spherical organelle responsible for destroying its contents . unlike the lysosome , which mostly degrades proteins , the peroxisome is the site of fatty acid breakdown . it also protects the cell from reactive oxygen species ( ros ) molecules which could seriously damage the cell . ross are molecules like oxygen ions or peroxides that are created as a byproduct of normal cellular metabolism , but also by radiation , tobacco , and drugs . they cause what is known as oxidative stress in the cell by reacting with and damaging dna and lipid-based molecules like cell membranes . these ross are the reason we need antioxidants in our diet . mitochondria just like a factory can ’ t run without electricity , a cell can ’ t run without energy . atp ( adenosine triphosphate ) is the energy currency of the cell , and is produced in a process known as cellular respiration . though the process begins in the cytoplasm , the bulk of the energy produced comes from later steps that take place in the mitochondria . like we saw with the nuclear envelope , there are actually two lipid bilayers that separate the mitochondrial contents from the cytoplasm . we refer to them as the inner and outer mitochondrial membranes . if we cross both membranes we end up in the matrix , where pyruvate is sent after it is created from the breakdown of glucose ( this is step 1 of cellular respiration , known as glycolysis ) .the space between the two membranes is called the intermembrane space , and it has a low ph ( is acidic ) because the electron transport chain embedded in the inner membrane pumps protons ( h+ ) into it . energy to make atp comes from protons moving back into the matrix down their gradient from the intermembrane space . mitochondria are also somewhat unique in that they are self-replicating and have their own dna , almost as if they were a completely separate cell . the prevailing theory , known as the endosymbiotic theory , is that eukaryotes were first formed by large prokaryotic cells engulfing smaller cells that looked a lot like mitochondria ( and chloroplasts , more on them later ) . instead of being digested , the engulfed cells remained intact and the arrangement turned out to be advantageous to both cells , which created a symbiotic relationship . so far we ’ ve discussed organelles , the membrane-bound structures within a cell that have some sort of specialized function . now let ’ s take a moment to talk about the scaffolding that ’ s holding all of this in place - the walls and beams of our factory . cytoskeleton within the cytoplasm there is network of protein fibers known as the cytoskeleton . this structure is responsible for both cell movement and stability . the major components of the cytoskeleton are microtubules , intermediate filaments , and microfilaments . microtubules microtubules are small tubes made from the protein tubulin . these tubules are found in cilia and flagella , structures involved in cell movement . they also help provide pathways for secretory vesicles to move through the cell , and are even involved in cell division as they are a part of the mitotic spindle , which pulls homologous chromosomes apart . intermediate filaments smaller than the microtubules , but larger than the microfilaments , the intermediate filaments are made of a variety of proteins such as keratin and/or neurofilament . they are very stable , and help provide structure to the nuclear envelope and anchor organelles . microfilaments microfilaments are the thinnest part of the cytoskeleton , and are made of actin [ a highly-conserved protein that is actually the most abundant protein in most eukaryotic cells ] . actin is both flexible and strong , making it a useful protein in cell movement . in the heart , contraction is mediated through an actin-myosin system . plants and platelets so far we ’ ve covered basic organelles found in a eukaryotic cell . however , not every cell has each of these organelles , and some cells have organelles we haven ’ t discussed . for example , plant cells have chloroplasts , organelles that resemble mitochondria and are responsible for turning sunlight into useful energy for the cell ( this is like factories that are powered by energy they collect via solar panels ) . on the other hand , platelets , blood cells responsible for clotting , have no nucleus and are in fact just fragments of cytoplasm contained within a cell membrane . eukaryotes vs bacteria vs archaea it is also important to keep in mind that organelles are found only in eukaryotes , one of the three major cell divisions . the other two major divisions , bacteria and archaea are known as prokaryotes , and have no membrane bound organelles within . consider the following : some diseases can be traced back to organelle lack / malformation . for example , inclusion-cell ( i-cell ) disease occurs due to a defect in the golgi . in order to mark enzymes that should be sent to lysosomes to help degrade unwanted molecules , the golgi has to bind them with a mannose 6-phosphate tag , like a shipping label . however , in patients with i-cell disease , one of the proteins that make this tag is mutated , and can not do its job , like a broken label machine . this means that proteins can not be targeted to lysosomes . these untagged proteins are the enzymes that are responsible for chopping up other proteins . what happens is the inactivated enzymes end up being sent outside the cell , while lysosomes clog up with undigested material . this disease is congenital , and usually fatal before patients reach 7 years of age . an interesting idea is that mitochondria can be used to trace maternal ancestry . since mitochondria are self-replicating and have their own dna , they are not determined by the genes found in the nucleus . instead , your mitochondria have developed from the mitochondria present in the female ovum ( egg ) that you developed from . defects in mitochondrial dna cause hereditary diseases that pass only from mother to children .
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all cells have membranes ( the building ) , dna ( the various blueprints ) , and ribosomes ( the production line ) , and so are able to make proteins ( the product - let ’ s say we ’ re making toys ) . this article will focus on eukaryotes , since they are the cell type that contains organelles . what ’ s found inside a cell an organelle ( think of it as a cell ’ s internal organ ) is a membrane bound structure found within a cell . just like cells have membranes to hold everything in , these mini-organs are also bound in a double layer of phospholipids to insulate their little compartments within the larger cells .
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why does the cell need a specific organelle just for breaking down nucleic acids ?
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what is a cell right now your body is doing a million things at once . it ’ s sending electrical impulses , pumping blood , filtering urine , digesting food , making protein , storing fat , and that ’ s just the stuff you ’ re not thinking about ! you can do all this because you are made of cells — tiny units of life that are like specialized factories , full of machinery designed to accomplish the business of life . cells make up every living thing , from blue whales to the archaebacteria that live inside volcanos . just like the organisms they make up , cells can come in all shapes and sizes . nerve cells in giant squids can reach up to 12m [ 39 ft ] in length , while human eggs ( the largest human cells ) are about 0.1mm across . plant cells have protective walls made of cellulose ( which also makes up the strings in celery that make it so hard to eat ) while fungal cell walls are made from the same stuff as lobster shells . however , despite this vast range in size , shape , and function , all these little factories have the same basic machinery . there are two main types of cells , prokaryotic and eukaryotic . prokaryotes are cells that do not have membrane bound nuclei , whereas eukaryotes do . the rest of our discussion will strictly be on eukaryotes . think about what a factory needs in order to function effectively . at its most basic , a factory needs a building , a product , and a way to make that product . all cells have membranes ( the building ) , dna ( the various blueprints ) , and ribosomes ( the production line ) , and so are able to make proteins ( the product - let ’ s say we ’ re making toys ) . this article will focus on eukaryotes , since they are the cell type that contains organelles . what ’ s found inside a cell an organelle ( think of it as a cell ’ s internal organ ) is a membrane bound structure found within a cell . just like cells have membranes to hold everything in , these mini-organs are also bound in a double layer of phospholipids to insulate their little compartments within the larger cells . you can think of organelles as smaller rooms within the factory , with specialized conditions to help these rooms carry out their specific task ( like a break room stocked with goodies or a research room with cool gadgets and a special air filter ) . these organelles are found in the cytoplasm , a viscous liquid found within the cell membrane that houses the organelles and is the location of most of the action happening in a cell . below is a table of the organelles found in the basic human cell , which we ’ ll be using as our template for this discussion . organelle | function | factory part : - : | : - : | : - : nucleus | dna storage | room where the blueprints are kept mitochondrion | energy production | powerplant smooth endoplasmic reticulum ( ser ) | lipid production ; detoxification | accessory production - makes decorations for the toy , etc . rough endoplasmic reticulum ( rer ) | protein production ; in particular for export out of the cell | primary production line - makes the toys golgi apparatus | protein modification and export | shipping department peroxisome | lipid destruction ; contains oxidative enzymes | security and waste removal lysosome | protein destruction | recycling and security nucleus our dna has the blueprints for every protein in our body , all packaged into a neat double helix . the processes to transform dna into proteins are known as transcription and translation , and happen in different compartments within the cell . the first step , transcription , happens in the nucleus , which holds our dna . a membrane called the nuclear envelope surrounds the nucleus , and its job is to create a room within the cell to both protect the genetic information and to house all the molecules that are involved in processing and protecting that info . this membrane is actually a set of two lipid bilayers , so there are four sheets of lipids separating the inside of the nucleus from the cytoplasm . the space between the two bilayers is known as the perinuclear space . though part of the function of the nucleus is to separate the dna from the rest of the cell , molecules must still be able to move in and out ( e.g. , rna ) . proteins channels known as nuclear pores form holes in the nuclear envelope . the nucleus itself is filled with liquid ( called nucleoplasm ) and is similar in structure and function to cytoplasm . it is here within the nucleoplasm where chromosomes ( tightly packed strands of dna containing all our blueprints ) are found . a nucleus has interesting implications for how a cell responds to its environment . thanks to the added protection of the nuclear envelope , the dna is a little bit more secure from enzymes , pathogens , and potentially harmful products of fat and protein metabolism . since this is the only permanent copy of the instructions the cell has , it is very important to keep the dna in good condition . if the dna was not sequestered away , it would be vulnerable to damage by the aforementioned dangers , which would then lead to defective protein production . imagine a giant hole or coffee stain in the blueprint for your toy - all of a sudden you don ’ t have either enough or the right information to make a critical piece of the toy . the nuclear envelope also keeps molecules responsible for dna transcription and repair close to the dna itself - otherwise those molecules would diffuse across the entire cell and it would take a lot more work and luck to get anything done ! while transcription ( making a complementary strand of rna from dna ) is completed within the nucleus , translation ( making protein from rna instructions ) takes place in the cytoplasm . if there was no barrier between the transcription and translation machineries , it ’ s possible that poorly-made or unfinished rna would get turned into poorly made and potentially dangerous proteins . before an rna can exit the nucleus to be translated , it must get special modifications , in the form of a cap and tail at either end of the molecule , that act as a stamp of approval to let the cell know this piece of rna is complete and properly made . nucleolus within the nucleus is a small subspace known as the nucleolus . it is not bound by a membrane , so it is not an organelle . this space forms near the part of dna with instructions for making ribosomes , the molecules responsible for making proteins . ribosomes are assembled in the nucleolus , and exit the nucleus with nuclear pores . in our analogy , the robots making our product are made in a special corner of the blueprint room , before being released to the factory . endoplasmic reticulum endoplasmic means inside ( endo ) the cytoplasm ( plasm ) . reticulum comes from the latin word for net . basically , an endoplasmic reticulum is a plasma membrane found inside the cell that folds in on itself to create an internal space known as the lumen . this lumen is actually continuous with the perinuclear space , so we know the endoplasmic reticulum is attached to the nuclear envelope . there are actually two different endoplasmic reticuli in a cell : the smooth endoplasmic reticulum and the rough endoplasmic reticulum . the rough endoplasmic reticulum is the site of protein production ( where we make our major product - the toy ) while the smooth endoplasmic reticulum is where lipids ( fats ) are made ( accessories for the toy , but not the central product of the factory ) . rough endoplasmic reticulum the rough endoplasmic reticulum is so-called because its surface is studded with ribosomes , the molecules in charge of protein production . when a ribosome finds a specific rna segment , that segment may tell the ribosome to travel to the rough endoplasmic reticulum and embed itself . the protein created from this segment will find itself inside the lumen of the rough endoplasmic reticulum , where it folds and is tagged with a ( usually carbohydrate ) molecule in a process known as glycosylation that marks the protein for transport to the golgi apparatus . the rough endoplasmic reticulum is continuous with the nuclear envelope , and looks like a series of canals near the nucleus . proteins made in the rough endoplasmic reticulum as destined to either be a part of a membrane , or to be secreted from the cell membrane out of the cell . without an rough endoplasmic reticulum , it would be a lot harder to distinguish between proteins that should leave the cell , and proteins that should remain . thus , the rough endoplasmic reticulum helps cells specialize and allows for greater complexity in the organism . smooth endoplasmic reticulum the smooth endoplasmic reticulum makes lipids and steroids , instead of being involved in protein synthesis . these are fat-based molecules that are important in energy storage , membrane structure , and communication ( steroids can act as hormones ) . the smooth endoplasmic reticulum is also responsible for detoxifying the cell . it is more tubular than the rough endoplasmic reticulum , and is not necessarily continuous with the nuclear envelope . every cell has a smooth endoplasmic reticulum , but the amount will vary with cell function . for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum . golgi apparatus ( aka golgi body aka golgi ) we mentioned the golgi apparatus earlier when we discussed the production of proteins in the rough endoplasmic reticulum . if the smooth and rough endoplasmic reticula are how we make our product , the golgi is the mailroom that sends our product to customers . it is responsible for packing proteins from the rough endoplasmic reticulum into membrane-bound vesicles ( tiny compartments of lipid bilayer that store molecules ) which then translocate to the cell membrane . at the cell membrane , the vesicles can fuse with the larger lipid bilayer , causing the vesicle contents to either become part of the cell membrane or be released to the outside . different molecules actually have different fates upon entering the golgi . this determination is done by tagging the proteins with special sugar molecules that act as a shipping label for the protein . the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it . molecules in this pathway are often protein channels which allow molecules into or out of the cell , or cell identifiers which project into the extracellular space and act like a name tag for the cell . secretion : some proteins are meant to be secreted from the cell to act on other parts of the body . before these vesicles can fuse with the cell membrane , they must accumulate in number , and require a special chemical signal to be released . this way shipments only go out if they ’ re worth the cost of sending them ( you generally wouldn ’ t ship just one toy and expect to profit ) . lysosome : the final destination for proteins coming through the golgi is the lysosome . vesicles sent to this acidic organelle contain enzymes that will hydrolyze the lysosome ’ s content . lysosome the lysosome is the cell ’ s recycling center . these organelles are spheres full of enzymes ready to hydrolyze ( chop up the chemical bonds of ) whatever substance crosses the membrane , so the cell can reuse the raw material . these disposal enzymes only function properly in environments with a ph of 5 , two orders of magnitude more acidic than the cell ’ s internal ph of 7 . lysosomal proteins only being active in an acidic environment acts as safety mechanism for the rest of the cell - if the lysosome were to somehow leak or burst , the degradative enzymes would inactivate before they chopped up proteins the cell still needed . peroxisome like the lysosome , the peroxisome is a spherical organelle responsible for destroying its contents . unlike the lysosome , which mostly degrades proteins , the peroxisome is the site of fatty acid breakdown . it also protects the cell from reactive oxygen species ( ros ) molecules which could seriously damage the cell . ross are molecules like oxygen ions or peroxides that are created as a byproduct of normal cellular metabolism , but also by radiation , tobacco , and drugs . they cause what is known as oxidative stress in the cell by reacting with and damaging dna and lipid-based molecules like cell membranes . these ross are the reason we need antioxidants in our diet . mitochondria just like a factory can ’ t run without electricity , a cell can ’ t run without energy . atp ( adenosine triphosphate ) is the energy currency of the cell , and is produced in a process known as cellular respiration . though the process begins in the cytoplasm , the bulk of the energy produced comes from later steps that take place in the mitochondria . like we saw with the nuclear envelope , there are actually two lipid bilayers that separate the mitochondrial contents from the cytoplasm . we refer to them as the inner and outer mitochondrial membranes . if we cross both membranes we end up in the matrix , where pyruvate is sent after it is created from the breakdown of glucose ( this is step 1 of cellular respiration , known as glycolysis ) .the space between the two membranes is called the intermembrane space , and it has a low ph ( is acidic ) because the electron transport chain embedded in the inner membrane pumps protons ( h+ ) into it . energy to make atp comes from protons moving back into the matrix down their gradient from the intermembrane space . mitochondria are also somewhat unique in that they are self-replicating and have their own dna , almost as if they were a completely separate cell . the prevailing theory , known as the endosymbiotic theory , is that eukaryotes were first formed by large prokaryotic cells engulfing smaller cells that looked a lot like mitochondria ( and chloroplasts , more on them later ) . instead of being digested , the engulfed cells remained intact and the arrangement turned out to be advantageous to both cells , which created a symbiotic relationship . so far we ’ ve discussed organelles , the membrane-bound structures within a cell that have some sort of specialized function . now let ’ s take a moment to talk about the scaffolding that ’ s holding all of this in place - the walls and beams of our factory . cytoskeleton within the cytoplasm there is network of protein fibers known as the cytoskeleton . this structure is responsible for both cell movement and stability . the major components of the cytoskeleton are microtubules , intermediate filaments , and microfilaments . microtubules microtubules are small tubes made from the protein tubulin . these tubules are found in cilia and flagella , structures involved in cell movement . they also help provide pathways for secretory vesicles to move through the cell , and are even involved in cell division as they are a part of the mitotic spindle , which pulls homologous chromosomes apart . intermediate filaments smaller than the microtubules , but larger than the microfilaments , the intermediate filaments are made of a variety of proteins such as keratin and/or neurofilament . they are very stable , and help provide structure to the nuclear envelope and anchor organelles . microfilaments microfilaments are the thinnest part of the cytoskeleton , and are made of actin [ a highly-conserved protein that is actually the most abundant protein in most eukaryotic cells ] . actin is both flexible and strong , making it a useful protein in cell movement . in the heart , contraction is mediated through an actin-myosin system . plants and platelets so far we ’ ve covered basic organelles found in a eukaryotic cell . however , not every cell has each of these organelles , and some cells have organelles we haven ’ t discussed . for example , plant cells have chloroplasts , organelles that resemble mitochondria and are responsible for turning sunlight into useful energy for the cell ( this is like factories that are powered by energy they collect via solar panels ) . on the other hand , platelets , blood cells responsible for clotting , have no nucleus and are in fact just fragments of cytoplasm contained within a cell membrane . eukaryotes vs bacteria vs archaea it is also important to keep in mind that organelles are found only in eukaryotes , one of the three major cell divisions . the other two major divisions , bacteria and archaea are known as prokaryotes , and have no membrane bound organelles within . consider the following : some diseases can be traced back to organelle lack / malformation . for example , inclusion-cell ( i-cell ) disease occurs due to a defect in the golgi . in order to mark enzymes that should be sent to lysosomes to help degrade unwanted molecules , the golgi has to bind them with a mannose 6-phosphate tag , like a shipping label . however , in patients with i-cell disease , one of the proteins that make this tag is mutated , and can not do its job , like a broken label machine . this means that proteins can not be targeted to lysosomes . these untagged proteins are the enzymes that are responsible for chopping up other proteins . what happens is the inactivated enzymes end up being sent outside the cell , while lysosomes clog up with undigested material . this disease is congenital , and usually fatal before patients reach 7 years of age . an interesting idea is that mitochondria can be used to trace maternal ancestry . since mitochondria are self-replicating and have their own dna , they are not determined by the genes found in the nucleus . instead , your mitochondria have developed from the mitochondria present in the female ovum ( egg ) that you developed from . defects in mitochondrial dna cause hereditary diseases that pass only from mother to children .
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for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum . golgi apparatus ( aka golgi body aka golgi ) we mentioned the golgi apparatus earlier when we discussed the production of proteins in the rough endoplasmic reticulum . if the smooth and rough endoplasmic reticula are how we make our product , the golgi is the mailroom that sends our product to customers .
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does it mean the golgi adds the tag to the protein , does it mean the golgi has the mannose 6-phosphate tag on it and it binds the protein using the tag , or does it mean the protein already has the tag and the golgi binds the tag ?
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what is a cell right now your body is doing a million things at once . it ’ s sending electrical impulses , pumping blood , filtering urine , digesting food , making protein , storing fat , and that ’ s just the stuff you ’ re not thinking about ! you can do all this because you are made of cells — tiny units of life that are like specialized factories , full of machinery designed to accomplish the business of life . cells make up every living thing , from blue whales to the archaebacteria that live inside volcanos . just like the organisms they make up , cells can come in all shapes and sizes . nerve cells in giant squids can reach up to 12m [ 39 ft ] in length , while human eggs ( the largest human cells ) are about 0.1mm across . plant cells have protective walls made of cellulose ( which also makes up the strings in celery that make it so hard to eat ) while fungal cell walls are made from the same stuff as lobster shells . however , despite this vast range in size , shape , and function , all these little factories have the same basic machinery . there are two main types of cells , prokaryotic and eukaryotic . prokaryotes are cells that do not have membrane bound nuclei , whereas eukaryotes do . the rest of our discussion will strictly be on eukaryotes . think about what a factory needs in order to function effectively . at its most basic , a factory needs a building , a product , and a way to make that product . all cells have membranes ( the building ) , dna ( the various blueprints ) , and ribosomes ( the production line ) , and so are able to make proteins ( the product - let ’ s say we ’ re making toys ) . this article will focus on eukaryotes , since they are the cell type that contains organelles . what ’ s found inside a cell an organelle ( think of it as a cell ’ s internal organ ) is a membrane bound structure found within a cell . just like cells have membranes to hold everything in , these mini-organs are also bound in a double layer of phospholipids to insulate their little compartments within the larger cells . you can think of organelles as smaller rooms within the factory , with specialized conditions to help these rooms carry out their specific task ( like a break room stocked with goodies or a research room with cool gadgets and a special air filter ) . these organelles are found in the cytoplasm , a viscous liquid found within the cell membrane that houses the organelles and is the location of most of the action happening in a cell . below is a table of the organelles found in the basic human cell , which we ’ ll be using as our template for this discussion . organelle | function | factory part : - : | : - : | : - : nucleus | dna storage | room where the blueprints are kept mitochondrion | energy production | powerplant smooth endoplasmic reticulum ( ser ) | lipid production ; detoxification | accessory production - makes decorations for the toy , etc . rough endoplasmic reticulum ( rer ) | protein production ; in particular for export out of the cell | primary production line - makes the toys golgi apparatus | protein modification and export | shipping department peroxisome | lipid destruction ; contains oxidative enzymes | security and waste removal lysosome | protein destruction | recycling and security nucleus our dna has the blueprints for every protein in our body , all packaged into a neat double helix . the processes to transform dna into proteins are known as transcription and translation , and happen in different compartments within the cell . the first step , transcription , happens in the nucleus , which holds our dna . a membrane called the nuclear envelope surrounds the nucleus , and its job is to create a room within the cell to both protect the genetic information and to house all the molecules that are involved in processing and protecting that info . this membrane is actually a set of two lipid bilayers , so there are four sheets of lipids separating the inside of the nucleus from the cytoplasm . the space between the two bilayers is known as the perinuclear space . though part of the function of the nucleus is to separate the dna from the rest of the cell , molecules must still be able to move in and out ( e.g. , rna ) . proteins channels known as nuclear pores form holes in the nuclear envelope . the nucleus itself is filled with liquid ( called nucleoplasm ) and is similar in structure and function to cytoplasm . it is here within the nucleoplasm where chromosomes ( tightly packed strands of dna containing all our blueprints ) are found . a nucleus has interesting implications for how a cell responds to its environment . thanks to the added protection of the nuclear envelope , the dna is a little bit more secure from enzymes , pathogens , and potentially harmful products of fat and protein metabolism . since this is the only permanent copy of the instructions the cell has , it is very important to keep the dna in good condition . if the dna was not sequestered away , it would be vulnerable to damage by the aforementioned dangers , which would then lead to defective protein production . imagine a giant hole or coffee stain in the blueprint for your toy - all of a sudden you don ’ t have either enough or the right information to make a critical piece of the toy . the nuclear envelope also keeps molecules responsible for dna transcription and repair close to the dna itself - otherwise those molecules would diffuse across the entire cell and it would take a lot more work and luck to get anything done ! while transcription ( making a complementary strand of rna from dna ) is completed within the nucleus , translation ( making protein from rna instructions ) takes place in the cytoplasm . if there was no barrier between the transcription and translation machineries , it ’ s possible that poorly-made or unfinished rna would get turned into poorly made and potentially dangerous proteins . before an rna can exit the nucleus to be translated , it must get special modifications , in the form of a cap and tail at either end of the molecule , that act as a stamp of approval to let the cell know this piece of rna is complete and properly made . nucleolus within the nucleus is a small subspace known as the nucleolus . it is not bound by a membrane , so it is not an organelle . this space forms near the part of dna with instructions for making ribosomes , the molecules responsible for making proteins . ribosomes are assembled in the nucleolus , and exit the nucleus with nuclear pores . in our analogy , the robots making our product are made in a special corner of the blueprint room , before being released to the factory . endoplasmic reticulum endoplasmic means inside ( endo ) the cytoplasm ( plasm ) . reticulum comes from the latin word for net . basically , an endoplasmic reticulum is a plasma membrane found inside the cell that folds in on itself to create an internal space known as the lumen . this lumen is actually continuous with the perinuclear space , so we know the endoplasmic reticulum is attached to the nuclear envelope . there are actually two different endoplasmic reticuli in a cell : the smooth endoplasmic reticulum and the rough endoplasmic reticulum . the rough endoplasmic reticulum is the site of protein production ( where we make our major product - the toy ) while the smooth endoplasmic reticulum is where lipids ( fats ) are made ( accessories for the toy , but not the central product of the factory ) . rough endoplasmic reticulum the rough endoplasmic reticulum is so-called because its surface is studded with ribosomes , the molecules in charge of protein production . when a ribosome finds a specific rna segment , that segment may tell the ribosome to travel to the rough endoplasmic reticulum and embed itself . the protein created from this segment will find itself inside the lumen of the rough endoplasmic reticulum , where it folds and is tagged with a ( usually carbohydrate ) molecule in a process known as glycosylation that marks the protein for transport to the golgi apparatus . the rough endoplasmic reticulum is continuous with the nuclear envelope , and looks like a series of canals near the nucleus . proteins made in the rough endoplasmic reticulum as destined to either be a part of a membrane , or to be secreted from the cell membrane out of the cell . without an rough endoplasmic reticulum , it would be a lot harder to distinguish between proteins that should leave the cell , and proteins that should remain . thus , the rough endoplasmic reticulum helps cells specialize and allows for greater complexity in the organism . smooth endoplasmic reticulum the smooth endoplasmic reticulum makes lipids and steroids , instead of being involved in protein synthesis . these are fat-based molecules that are important in energy storage , membrane structure , and communication ( steroids can act as hormones ) . the smooth endoplasmic reticulum is also responsible for detoxifying the cell . it is more tubular than the rough endoplasmic reticulum , and is not necessarily continuous with the nuclear envelope . every cell has a smooth endoplasmic reticulum , but the amount will vary with cell function . for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum . golgi apparatus ( aka golgi body aka golgi ) we mentioned the golgi apparatus earlier when we discussed the production of proteins in the rough endoplasmic reticulum . if the smooth and rough endoplasmic reticula are how we make our product , the golgi is the mailroom that sends our product to customers . it is responsible for packing proteins from the rough endoplasmic reticulum into membrane-bound vesicles ( tiny compartments of lipid bilayer that store molecules ) which then translocate to the cell membrane . at the cell membrane , the vesicles can fuse with the larger lipid bilayer , causing the vesicle contents to either become part of the cell membrane or be released to the outside . different molecules actually have different fates upon entering the golgi . this determination is done by tagging the proteins with special sugar molecules that act as a shipping label for the protein . the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it . molecules in this pathway are often protein channels which allow molecules into or out of the cell , or cell identifiers which project into the extracellular space and act like a name tag for the cell . secretion : some proteins are meant to be secreted from the cell to act on other parts of the body . before these vesicles can fuse with the cell membrane , they must accumulate in number , and require a special chemical signal to be released . this way shipments only go out if they ’ re worth the cost of sending them ( you generally wouldn ’ t ship just one toy and expect to profit ) . lysosome : the final destination for proteins coming through the golgi is the lysosome . vesicles sent to this acidic organelle contain enzymes that will hydrolyze the lysosome ’ s content . lysosome the lysosome is the cell ’ s recycling center . these organelles are spheres full of enzymes ready to hydrolyze ( chop up the chemical bonds of ) whatever substance crosses the membrane , so the cell can reuse the raw material . these disposal enzymes only function properly in environments with a ph of 5 , two orders of magnitude more acidic than the cell ’ s internal ph of 7 . lysosomal proteins only being active in an acidic environment acts as safety mechanism for the rest of the cell - if the lysosome were to somehow leak or burst , the degradative enzymes would inactivate before they chopped up proteins the cell still needed . peroxisome like the lysosome , the peroxisome is a spherical organelle responsible for destroying its contents . unlike the lysosome , which mostly degrades proteins , the peroxisome is the site of fatty acid breakdown . it also protects the cell from reactive oxygen species ( ros ) molecules which could seriously damage the cell . ross are molecules like oxygen ions or peroxides that are created as a byproduct of normal cellular metabolism , but also by radiation , tobacco , and drugs . they cause what is known as oxidative stress in the cell by reacting with and damaging dna and lipid-based molecules like cell membranes . these ross are the reason we need antioxidants in our diet . mitochondria just like a factory can ’ t run without electricity , a cell can ’ t run without energy . atp ( adenosine triphosphate ) is the energy currency of the cell , and is produced in a process known as cellular respiration . though the process begins in the cytoplasm , the bulk of the energy produced comes from later steps that take place in the mitochondria . like we saw with the nuclear envelope , there are actually two lipid bilayers that separate the mitochondrial contents from the cytoplasm . we refer to them as the inner and outer mitochondrial membranes . if we cross both membranes we end up in the matrix , where pyruvate is sent after it is created from the breakdown of glucose ( this is step 1 of cellular respiration , known as glycolysis ) .the space between the two membranes is called the intermembrane space , and it has a low ph ( is acidic ) because the electron transport chain embedded in the inner membrane pumps protons ( h+ ) into it . energy to make atp comes from protons moving back into the matrix down their gradient from the intermembrane space . mitochondria are also somewhat unique in that they are self-replicating and have their own dna , almost as if they were a completely separate cell . the prevailing theory , known as the endosymbiotic theory , is that eukaryotes were first formed by large prokaryotic cells engulfing smaller cells that looked a lot like mitochondria ( and chloroplasts , more on them later ) . instead of being digested , the engulfed cells remained intact and the arrangement turned out to be advantageous to both cells , which created a symbiotic relationship . so far we ’ ve discussed organelles , the membrane-bound structures within a cell that have some sort of specialized function . now let ’ s take a moment to talk about the scaffolding that ’ s holding all of this in place - the walls and beams of our factory . cytoskeleton within the cytoplasm there is network of protein fibers known as the cytoskeleton . this structure is responsible for both cell movement and stability . the major components of the cytoskeleton are microtubules , intermediate filaments , and microfilaments . microtubules microtubules are small tubes made from the protein tubulin . these tubules are found in cilia and flagella , structures involved in cell movement . they also help provide pathways for secretory vesicles to move through the cell , and are even involved in cell division as they are a part of the mitotic spindle , which pulls homologous chromosomes apart . intermediate filaments smaller than the microtubules , but larger than the microfilaments , the intermediate filaments are made of a variety of proteins such as keratin and/or neurofilament . they are very stable , and help provide structure to the nuclear envelope and anchor organelles . microfilaments microfilaments are the thinnest part of the cytoskeleton , and are made of actin [ a highly-conserved protein that is actually the most abundant protein in most eukaryotic cells ] . actin is both flexible and strong , making it a useful protein in cell movement . in the heart , contraction is mediated through an actin-myosin system . plants and platelets so far we ’ ve covered basic organelles found in a eukaryotic cell . however , not every cell has each of these organelles , and some cells have organelles we haven ’ t discussed . for example , plant cells have chloroplasts , organelles that resemble mitochondria and are responsible for turning sunlight into useful energy for the cell ( this is like factories that are powered by energy they collect via solar panels ) . on the other hand , platelets , blood cells responsible for clotting , have no nucleus and are in fact just fragments of cytoplasm contained within a cell membrane . eukaryotes vs bacteria vs archaea it is also important to keep in mind that organelles are found only in eukaryotes , one of the three major cell divisions . the other two major divisions , bacteria and archaea are known as prokaryotes , and have no membrane bound organelles within . consider the following : some diseases can be traced back to organelle lack / malformation . for example , inclusion-cell ( i-cell ) disease occurs due to a defect in the golgi . in order to mark enzymes that should be sent to lysosomes to help degrade unwanted molecules , the golgi has to bind them with a mannose 6-phosphate tag , like a shipping label . however , in patients with i-cell disease , one of the proteins that make this tag is mutated , and can not do its job , like a broken label machine . this means that proteins can not be targeted to lysosomes . these untagged proteins are the enzymes that are responsible for chopping up other proteins . what happens is the inactivated enzymes end up being sent outside the cell , while lysosomes clog up with undigested material . this disease is congenital , and usually fatal before patients reach 7 years of age . an interesting idea is that mitochondria can be used to trace maternal ancestry . since mitochondria are self-replicating and have their own dna , they are not determined by the genes found in the nucleus . instead , your mitochondria have developed from the mitochondria present in the female ovum ( egg ) that you developed from . defects in mitochondrial dna cause hereditary diseases that pass only from mother to children .
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it is more tubular than the rough endoplasmic reticulum , and is not necessarily continuous with the nuclear envelope . every cell has a smooth endoplasmic reticulum , but the amount will vary with cell function . for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum .
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how does the smooth er detoxify the cell ?
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what is a cell right now your body is doing a million things at once . it ’ s sending electrical impulses , pumping blood , filtering urine , digesting food , making protein , storing fat , and that ’ s just the stuff you ’ re not thinking about ! you can do all this because you are made of cells — tiny units of life that are like specialized factories , full of machinery designed to accomplish the business of life . cells make up every living thing , from blue whales to the archaebacteria that live inside volcanos . just like the organisms they make up , cells can come in all shapes and sizes . nerve cells in giant squids can reach up to 12m [ 39 ft ] in length , while human eggs ( the largest human cells ) are about 0.1mm across . plant cells have protective walls made of cellulose ( which also makes up the strings in celery that make it so hard to eat ) while fungal cell walls are made from the same stuff as lobster shells . however , despite this vast range in size , shape , and function , all these little factories have the same basic machinery . there are two main types of cells , prokaryotic and eukaryotic . prokaryotes are cells that do not have membrane bound nuclei , whereas eukaryotes do . the rest of our discussion will strictly be on eukaryotes . think about what a factory needs in order to function effectively . at its most basic , a factory needs a building , a product , and a way to make that product . all cells have membranes ( the building ) , dna ( the various blueprints ) , and ribosomes ( the production line ) , and so are able to make proteins ( the product - let ’ s say we ’ re making toys ) . this article will focus on eukaryotes , since they are the cell type that contains organelles . what ’ s found inside a cell an organelle ( think of it as a cell ’ s internal organ ) is a membrane bound structure found within a cell . just like cells have membranes to hold everything in , these mini-organs are also bound in a double layer of phospholipids to insulate their little compartments within the larger cells . you can think of organelles as smaller rooms within the factory , with specialized conditions to help these rooms carry out their specific task ( like a break room stocked with goodies or a research room with cool gadgets and a special air filter ) . these organelles are found in the cytoplasm , a viscous liquid found within the cell membrane that houses the organelles and is the location of most of the action happening in a cell . below is a table of the organelles found in the basic human cell , which we ’ ll be using as our template for this discussion . organelle | function | factory part : - : | : - : | : - : nucleus | dna storage | room where the blueprints are kept mitochondrion | energy production | powerplant smooth endoplasmic reticulum ( ser ) | lipid production ; detoxification | accessory production - makes decorations for the toy , etc . rough endoplasmic reticulum ( rer ) | protein production ; in particular for export out of the cell | primary production line - makes the toys golgi apparatus | protein modification and export | shipping department peroxisome | lipid destruction ; contains oxidative enzymes | security and waste removal lysosome | protein destruction | recycling and security nucleus our dna has the blueprints for every protein in our body , all packaged into a neat double helix . the processes to transform dna into proteins are known as transcription and translation , and happen in different compartments within the cell . the first step , transcription , happens in the nucleus , which holds our dna . a membrane called the nuclear envelope surrounds the nucleus , and its job is to create a room within the cell to both protect the genetic information and to house all the molecules that are involved in processing and protecting that info . this membrane is actually a set of two lipid bilayers , so there are four sheets of lipids separating the inside of the nucleus from the cytoplasm . the space between the two bilayers is known as the perinuclear space . though part of the function of the nucleus is to separate the dna from the rest of the cell , molecules must still be able to move in and out ( e.g. , rna ) . proteins channels known as nuclear pores form holes in the nuclear envelope . the nucleus itself is filled with liquid ( called nucleoplasm ) and is similar in structure and function to cytoplasm . it is here within the nucleoplasm where chromosomes ( tightly packed strands of dna containing all our blueprints ) are found . a nucleus has interesting implications for how a cell responds to its environment . thanks to the added protection of the nuclear envelope , the dna is a little bit more secure from enzymes , pathogens , and potentially harmful products of fat and protein metabolism . since this is the only permanent copy of the instructions the cell has , it is very important to keep the dna in good condition . if the dna was not sequestered away , it would be vulnerable to damage by the aforementioned dangers , which would then lead to defective protein production . imagine a giant hole or coffee stain in the blueprint for your toy - all of a sudden you don ’ t have either enough or the right information to make a critical piece of the toy . the nuclear envelope also keeps molecules responsible for dna transcription and repair close to the dna itself - otherwise those molecules would diffuse across the entire cell and it would take a lot more work and luck to get anything done ! while transcription ( making a complementary strand of rna from dna ) is completed within the nucleus , translation ( making protein from rna instructions ) takes place in the cytoplasm . if there was no barrier between the transcription and translation machineries , it ’ s possible that poorly-made or unfinished rna would get turned into poorly made and potentially dangerous proteins . before an rna can exit the nucleus to be translated , it must get special modifications , in the form of a cap and tail at either end of the molecule , that act as a stamp of approval to let the cell know this piece of rna is complete and properly made . nucleolus within the nucleus is a small subspace known as the nucleolus . it is not bound by a membrane , so it is not an organelle . this space forms near the part of dna with instructions for making ribosomes , the molecules responsible for making proteins . ribosomes are assembled in the nucleolus , and exit the nucleus with nuclear pores . in our analogy , the robots making our product are made in a special corner of the blueprint room , before being released to the factory . endoplasmic reticulum endoplasmic means inside ( endo ) the cytoplasm ( plasm ) . reticulum comes from the latin word for net . basically , an endoplasmic reticulum is a plasma membrane found inside the cell that folds in on itself to create an internal space known as the lumen . this lumen is actually continuous with the perinuclear space , so we know the endoplasmic reticulum is attached to the nuclear envelope . there are actually two different endoplasmic reticuli in a cell : the smooth endoplasmic reticulum and the rough endoplasmic reticulum . the rough endoplasmic reticulum is the site of protein production ( where we make our major product - the toy ) while the smooth endoplasmic reticulum is where lipids ( fats ) are made ( accessories for the toy , but not the central product of the factory ) . rough endoplasmic reticulum the rough endoplasmic reticulum is so-called because its surface is studded with ribosomes , the molecules in charge of protein production . when a ribosome finds a specific rna segment , that segment may tell the ribosome to travel to the rough endoplasmic reticulum and embed itself . the protein created from this segment will find itself inside the lumen of the rough endoplasmic reticulum , where it folds and is tagged with a ( usually carbohydrate ) molecule in a process known as glycosylation that marks the protein for transport to the golgi apparatus . the rough endoplasmic reticulum is continuous with the nuclear envelope , and looks like a series of canals near the nucleus . proteins made in the rough endoplasmic reticulum as destined to either be a part of a membrane , or to be secreted from the cell membrane out of the cell . without an rough endoplasmic reticulum , it would be a lot harder to distinguish between proteins that should leave the cell , and proteins that should remain . thus , the rough endoplasmic reticulum helps cells specialize and allows for greater complexity in the organism . smooth endoplasmic reticulum the smooth endoplasmic reticulum makes lipids and steroids , instead of being involved in protein synthesis . these are fat-based molecules that are important in energy storage , membrane structure , and communication ( steroids can act as hormones ) . the smooth endoplasmic reticulum is also responsible for detoxifying the cell . it is more tubular than the rough endoplasmic reticulum , and is not necessarily continuous with the nuclear envelope . every cell has a smooth endoplasmic reticulum , but the amount will vary with cell function . for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum . golgi apparatus ( aka golgi body aka golgi ) we mentioned the golgi apparatus earlier when we discussed the production of proteins in the rough endoplasmic reticulum . if the smooth and rough endoplasmic reticula are how we make our product , the golgi is the mailroom that sends our product to customers . it is responsible for packing proteins from the rough endoplasmic reticulum into membrane-bound vesicles ( tiny compartments of lipid bilayer that store molecules ) which then translocate to the cell membrane . at the cell membrane , the vesicles can fuse with the larger lipid bilayer , causing the vesicle contents to either become part of the cell membrane or be released to the outside . different molecules actually have different fates upon entering the golgi . this determination is done by tagging the proteins with special sugar molecules that act as a shipping label for the protein . the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it . molecules in this pathway are often protein channels which allow molecules into or out of the cell , or cell identifiers which project into the extracellular space and act like a name tag for the cell . secretion : some proteins are meant to be secreted from the cell to act on other parts of the body . before these vesicles can fuse with the cell membrane , they must accumulate in number , and require a special chemical signal to be released . this way shipments only go out if they ’ re worth the cost of sending them ( you generally wouldn ’ t ship just one toy and expect to profit ) . lysosome : the final destination for proteins coming through the golgi is the lysosome . vesicles sent to this acidic organelle contain enzymes that will hydrolyze the lysosome ’ s content . lysosome the lysosome is the cell ’ s recycling center . these organelles are spheres full of enzymes ready to hydrolyze ( chop up the chemical bonds of ) whatever substance crosses the membrane , so the cell can reuse the raw material . these disposal enzymes only function properly in environments with a ph of 5 , two orders of magnitude more acidic than the cell ’ s internal ph of 7 . lysosomal proteins only being active in an acidic environment acts as safety mechanism for the rest of the cell - if the lysosome were to somehow leak or burst , the degradative enzymes would inactivate before they chopped up proteins the cell still needed . peroxisome like the lysosome , the peroxisome is a spherical organelle responsible for destroying its contents . unlike the lysosome , which mostly degrades proteins , the peroxisome is the site of fatty acid breakdown . it also protects the cell from reactive oxygen species ( ros ) molecules which could seriously damage the cell . ross are molecules like oxygen ions or peroxides that are created as a byproduct of normal cellular metabolism , but also by radiation , tobacco , and drugs . they cause what is known as oxidative stress in the cell by reacting with and damaging dna and lipid-based molecules like cell membranes . these ross are the reason we need antioxidants in our diet . mitochondria just like a factory can ’ t run without electricity , a cell can ’ t run without energy . atp ( adenosine triphosphate ) is the energy currency of the cell , and is produced in a process known as cellular respiration . though the process begins in the cytoplasm , the bulk of the energy produced comes from later steps that take place in the mitochondria . like we saw with the nuclear envelope , there are actually two lipid bilayers that separate the mitochondrial contents from the cytoplasm . we refer to them as the inner and outer mitochondrial membranes . if we cross both membranes we end up in the matrix , where pyruvate is sent after it is created from the breakdown of glucose ( this is step 1 of cellular respiration , known as glycolysis ) .the space between the two membranes is called the intermembrane space , and it has a low ph ( is acidic ) because the electron transport chain embedded in the inner membrane pumps protons ( h+ ) into it . energy to make atp comes from protons moving back into the matrix down their gradient from the intermembrane space . mitochondria are also somewhat unique in that they are self-replicating and have their own dna , almost as if they were a completely separate cell . the prevailing theory , known as the endosymbiotic theory , is that eukaryotes were first formed by large prokaryotic cells engulfing smaller cells that looked a lot like mitochondria ( and chloroplasts , more on them later ) . instead of being digested , the engulfed cells remained intact and the arrangement turned out to be advantageous to both cells , which created a symbiotic relationship . so far we ’ ve discussed organelles , the membrane-bound structures within a cell that have some sort of specialized function . now let ’ s take a moment to talk about the scaffolding that ’ s holding all of this in place - the walls and beams of our factory . cytoskeleton within the cytoplasm there is network of protein fibers known as the cytoskeleton . this structure is responsible for both cell movement and stability . the major components of the cytoskeleton are microtubules , intermediate filaments , and microfilaments . microtubules microtubules are small tubes made from the protein tubulin . these tubules are found in cilia and flagella , structures involved in cell movement . they also help provide pathways for secretory vesicles to move through the cell , and are even involved in cell division as they are a part of the mitotic spindle , which pulls homologous chromosomes apart . intermediate filaments smaller than the microtubules , but larger than the microfilaments , the intermediate filaments are made of a variety of proteins such as keratin and/or neurofilament . they are very stable , and help provide structure to the nuclear envelope and anchor organelles . microfilaments microfilaments are the thinnest part of the cytoskeleton , and are made of actin [ a highly-conserved protein that is actually the most abundant protein in most eukaryotic cells ] . actin is both flexible and strong , making it a useful protein in cell movement . in the heart , contraction is mediated through an actin-myosin system . plants and platelets so far we ’ ve covered basic organelles found in a eukaryotic cell . however , not every cell has each of these organelles , and some cells have organelles we haven ’ t discussed . for example , plant cells have chloroplasts , organelles that resemble mitochondria and are responsible for turning sunlight into useful energy for the cell ( this is like factories that are powered by energy they collect via solar panels ) . on the other hand , platelets , blood cells responsible for clotting , have no nucleus and are in fact just fragments of cytoplasm contained within a cell membrane . eukaryotes vs bacteria vs archaea it is also important to keep in mind that organelles are found only in eukaryotes , one of the three major cell divisions . the other two major divisions , bacteria and archaea are known as prokaryotes , and have no membrane bound organelles within . consider the following : some diseases can be traced back to organelle lack / malformation . for example , inclusion-cell ( i-cell ) disease occurs due to a defect in the golgi . in order to mark enzymes that should be sent to lysosomes to help degrade unwanted molecules , the golgi has to bind them with a mannose 6-phosphate tag , like a shipping label . however , in patients with i-cell disease , one of the proteins that make this tag is mutated , and can not do its job , like a broken label machine . this means that proteins can not be targeted to lysosomes . these untagged proteins are the enzymes that are responsible for chopping up other proteins . what happens is the inactivated enzymes end up being sent outside the cell , while lysosomes clog up with undigested material . this disease is congenital , and usually fatal before patients reach 7 years of age . an interesting idea is that mitochondria can be used to trace maternal ancestry . since mitochondria are self-replicating and have their own dna , they are not determined by the genes found in the nucleus . instead , your mitochondria have developed from the mitochondria present in the female ovum ( egg ) that you developed from . defects in mitochondrial dna cause hereditary diseases that pass only from mother to children .
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the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it .
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what is an easy way to tell the difference between lysosomes and peroxisomes in a picture of a cell that is unlabeled ?
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what is a cell right now your body is doing a million things at once . it ’ s sending electrical impulses , pumping blood , filtering urine , digesting food , making protein , storing fat , and that ’ s just the stuff you ’ re not thinking about ! you can do all this because you are made of cells — tiny units of life that are like specialized factories , full of machinery designed to accomplish the business of life . cells make up every living thing , from blue whales to the archaebacteria that live inside volcanos . just like the organisms they make up , cells can come in all shapes and sizes . nerve cells in giant squids can reach up to 12m [ 39 ft ] in length , while human eggs ( the largest human cells ) are about 0.1mm across . plant cells have protective walls made of cellulose ( which also makes up the strings in celery that make it so hard to eat ) while fungal cell walls are made from the same stuff as lobster shells . however , despite this vast range in size , shape , and function , all these little factories have the same basic machinery . there are two main types of cells , prokaryotic and eukaryotic . prokaryotes are cells that do not have membrane bound nuclei , whereas eukaryotes do . the rest of our discussion will strictly be on eukaryotes . think about what a factory needs in order to function effectively . at its most basic , a factory needs a building , a product , and a way to make that product . all cells have membranes ( the building ) , dna ( the various blueprints ) , and ribosomes ( the production line ) , and so are able to make proteins ( the product - let ’ s say we ’ re making toys ) . this article will focus on eukaryotes , since they are the cell type that contains organelles . what ’ s found inside a cell an organelle ( think of it as a cell ’ s internal organ ) is a membrane bound structure found within a cell . just like cells have membranes to hold everything in , these mini-organs are also bound in a double layer of phospholipids to insulate their little compartments within the larger cells . you can think of organelles as smaller rooms within the factory , with specialized conditions to help these rooms carry out their specific task ( like a break room stocked with goodies or a research room with cool gadgets and a special air filter ) . these organelles are found in the cytoplasm , a viscous liquid found within the cell membrane that houses the organelles and is the location of most of the action happening in a cell . below is a table of the organelles found in the basic human cell , which we ’ ll be using as our template for this discussion . organelle | function | factory part : - : | : - : | : - : nucleus | dna storage | room where the blueprints are kept mitochondrion | energy production | powerplant smooth endoplasmic reticulum ( ser ) | lipid production ; detoxification | accessory production - makes decorations for the toy , etc . rough endoplasmic reticulum ( rer ) | protein production ; in particular for export out of the cell | primary production line - makes the toys golgi apparatus | protein modification and export | shipping department peroxisome | lipid destruction ; contains oxidative enzymes | security and waste removal lysosome | protein destruction | recycling and security nucleus our dna has the blueprints for every protein in our body , all packaged into a neat double helix . the processes to transform dna into proteins are known as transcription and translation , and happen in different compartments within the cell . the first step , transcription , happens in the nucleus , which holds our dna . a membrane called the nuclear envelope surrounds the nucleus , and its job is to create a room within the cell to both protect the genetic information and to house all the molecules that are involved in processing and protecting that info . this membrane is actually a set of two lipid bilayers , so there are four sheets of lipids separating the inside of the nucleus from the cytoplasm . the space between the two bilayers is known as the perinuclear space . though part of the function of the nucleus is to separate the dna from the rest of the cell , molecules must still be able to move in and out ( e.g. , rna ) . proteins channels known as nuclear pores form holes in the nuclear envelope . the nucleus itself is filled with liquid ( called nucleoplasm ) and is similar in structure and function to cytoplasm . it is here within the nucleoplasm where chromosomes ( tightly packed strands of dna containing all our blueprints ) are found . a nucleus has interesting implications for how a cell responds to its environment . thanks to the added protection of the nuclear envelope , the dna is a little bit more secure from enzymes , pathogens , and potentially harmful products of fat and protein metabolism . since this is the only permanent copy of the instructions the cell has , it is very important to keep the dna in good condition . if the dna was not sequestered away , it would be vulnerable to damage by the aforementioned dangers , which would then lead to defective protein production . imagine a giant hole or coffee stain in the blueprint for your toy - all of a sudden you don ’ t have either enough or the right information to make a critical piece of the toy . the nuclear envelope also keeps molecules responsible for dna transcription and repair close to the dna itself - otherwise those molecules would diffuse across the entire cell and it would take a lot more work and luck to get anything done ! while transcription ( making a complementary strand of rna from dna ) is completed within the nucleus , translation ( making protein from rna instructions ) takes place in the cytoplasm . if there was no barrier between the transcription and translation machineries , it ’ s possible that poorly-made or unfinished rna would get turned into poorly made and potentially dangerous proteins . before an rna can exit the nucleus to be translated , it must get special modifications , in the form of a cap and tail at either end of the molecule , that act as a stamp of approval to let the cell know this piece of rna is complete and properly made . nucleolus within the nucleus is a small subspace known as the nucleolus . it is not bound by a membrane , so it is not an organelle . this space forms near the part of dna with instructions for making ribosomes , the molecules responsible for making proteins . ribosomes are assembled in the nucleolus , and exit the nucleus with nuclear pores . in our analogy , the robots making our product are made in a special corner of the blueprint room , before being released to the factory . endoplasmic reticulum endoplasmic means inside ( endo ) the cytoplasm ( plasm ) . reticulum comes from the latin word for net . basically , an endoplasmic reticulum is a plasma membrane found inside the cell that folds in on itself to create an internal space known as the lumen . this lumen is actually continuous with the perinuclear space , so we know the endoplasmic reticulum is attached to the nuclear envelope . there are actually two different endoplasmic reticuli in a cell : the smooth endoplasmic reticulum and the rough endoplasmic reticulum . the rough endoplasmic reticulum is the site of protein production ( where we make our major product - the toy ) while the smooth endoplasmic reticulum is where lipids ( fats ) are made ( accessories for the toy , but not the central product of the factory ) . rough endoplasmic reticulum the rough endoplasmic reticulum is so-called because its surface is studded with ribosomes , the molecules in charge of protein production . when a ribosome finds a specific rna segment , that segment may tell the ribosome to travel to the rough endoplasmic reticulum and embed itself . the protein created from this segment will find itself inside the lumen of the rough endoplasmic reticulum , where it folds and is tagged with a ( usually carbohydrate ) molecule in a process known as glycosylation that marks the protein for transport to the golgi apparatus . the rough endoplasmic reticulum is continuous with the nuclear envelope , and looks like a series of canals near the nucleus . proteins made in the rough endoplasmic reticulum as destined to either be a part of a membrane , or to be secreted from the cell membrane out of the cell . without an rough endoplasmic reticulum , it would be a lot harder to distinguish between proteins that should leave the cell , and proteins that should remain . thus , the rough endoplasmic reticulum helps cells specialize and allows for greater complexity in the organism . smooth endoplasmic reticulum the smooth endoplasmic reticulum makes lipids and steroids , instead of being involved in protein synthesis . these are fat-based molecules that are important in energy storage , membrane structure , and communication ( steroids can act as hormones ) . the smooth endoplasmic reticulum is also responsible for detoxifying the cell . it is more tubular than the rough endoplasmic reticulum , and is not necessarily continuous with the nuclear envelope . every cell has a smooth endoplasmic reticulum , but the amount will vary with cell function . for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum . golgi apparatus ( aka golgi body aka golgi ) we mentioned the golgi apparatus earlier when we discussed the production of proteins in the rough endoplasmic reticulum . if the smooth and rough endoplasmic reticula are how we make our product , the golgi is the mailroom that sends our product to customers . it is responsible for packing proteins from the rough endoplasmic reticulum into membrane-bound vesicles ( tiny compartments of lipid bilayer that store molecules ) which then translocate to the cell membrane . at the cell membrane , the vesicles can fuse with the larger lipid bilayer , causing the vesicle contents to either become part of the cell membrane or be released to the outside . different molecules actually have different fates upon entering the golgi . this determination is done by tagging the proteins with special sugar molecules that act as a shipping label for the protein . the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it . molecules in this pathway are often protein channels which allow molecules into or out of the cell , or cell identifiers which project into the extracellular space and act like a name tag for the cell . secretion : some proteins are meant to be secreted from the cell to act on other parts of the body . before these vesicles can fuse with the cell membrane , they must accumulate in number , and require a special chemical signal to be released . this way shipments only go out if they ’ re worth the cost of sending them ( you generally wouldn ’ t ship just one toy and expect to profit ) . lysosome : the final destination for proteins coming through the golgi is the lysosome . vesicles sent to this acidic organelle contain enzymes that will hydrolyze the lysosome ’ s content . lysosome the lysosome is the cell ’ s recycling center . these organelles are spheres full of enzymes ready to hydrolyze ( chop up the chemical bonds of ) whatever substance crosses the membrane , so the cell can reuse the raw material . these disposal enzymes only function properly in environments with a ph of 5 , two orders of magnitude more acidic than the cell ’ s internal ph of 7 . lysosomal proteins only being active in an acidic environment acts as safety mechanism for the rest of the cell - if the lysosome were to somehow leak or burst , the degradative enzymes would inactivate before they chopped up proteins the cell still needed . peroxisome like the lysosome , the peroxisome is a spherical organelle responsible for destroying its contents . unlike the lysosome , which mostly degrades proteins , the peroxisome is the site of fatty acid breakdown . it also protects the cell from reactive oxygen species ( ros ) molecules which could seriously damage the cell . ross are molecules like oxygen ions or peroxides that are created as a byproduct of normal cellular metabolism , but also by radiation , tobacco , and drugs . they cause what is known as oxidative stress in the cell by reacting with and damaging dna and lipid-based molecules like cell membranes . these ross are the reason we need antioxidants in our diet . mitochondria just like a factory can ’ t run without electricity , a cell can ’ t run without energy . atp ( adenosine triphosphate ) is the energy currency of the cell , and is produced in a process known as cellular respiration . though the process begins in the cytoplasm , the bulk of the energy produced comes from later steps that take place in the mitochondria . like we saw with the nuclear envelope , there are actually two lipid bilayers that separate the mitochondrial contents from the cytoplasm . we refer to them as the inner and outer mitochondrial membranes . if we cross both membranes we end up in the matrix , where pyruvate is sent after it is created from the breakdown of glucose ( this is step 1 of cellular respiration , known as glycolysis ) .the space between the two membranes is called the intermembrane space , and it has a low ph ( is acidic ) because the electron transport chain embedded in the inner membrane pumps protons ( h+ ) into it . energy to make atp comes from protons moving back into the matrix down their gradient from the intermembrane space . mitochondria are also somewhat unique in that they are self-replicating and have their own dna , almost as if they were a completely separate cell . the prevailing theory , known as the endosymbiotic theory , is that eukaryotes were first formed by large prokaryotic cells engulfing smaller cells that looked a lot like mitochondria ( and chloroplasts , more on them later ) . instead of being digested , the engulfed cells remained intact and the arrangement turned out to be advantageous to both cells , which created a symbiotic relationship . so far we ’ ve discussed organelles , the membrane-bound structures within a cell that have some sort of specialized function . now let ’ s take a moment to talk about the scaffolding that ’ s holding all of this in place - the walls and beams of our factory . cytoskeleton within the cytoplasm there is network of protein fibers known as the cytoskeleton . this structure is responsible for both cell movement and stability . the major components of the cytoskeleton are microtubules , intermediate filaments , and microfilaments . microtubules microtubules are small tubes made from the protein tubulin . these tubules are found in cilia and flagella , structures involved in cell movement . they also help provide pathways for secretory vesicles to move through the cell , and are even involved in cell division as they are a part of the mitotic spindle , which pulls homologous chromosomes apart . intermediate filaments smaller than the microtubules , but larger than the microfilaments , the intermediate filaments are made of a variety of proteins such as keratin and/or neurofilament . they are very stable , and help provide structure to the nuclear envelope and anchor organelles . microfilaments microfilaments are the thinnest part of the cytoskeleton , and are made of actin [ a highly-conserved protein that is actually the most abundant protein in most eukaryotic cells ] . actin is both flexible and strong , making it a useful protein in cell movement . in the heart , contraction is mediated through an actin-myosin system . plants and platelets so far we ’ ve covered basic organelles found in a eukaryotic cell . however , not every cell has each of these organelles , and some cells have organelles we haven ’ t discussed . for example , plant cells have chloroplasts , organelles that resemble mitochondria and are responsible for turning sunlight into useful energy for the cell ( this is like factories that are powered by energy they collect via solar panels ) . on the other hand , platelets , blood cells responsible for clotting , have no nucleus and are in fact just fragments of cytoplasm contained within a cell membrane . eukaryotes vs bacteria vs archaea it is also important to keep in mind that organelles are found only in eukaryotes , one of the three major cell divisions . the other two major divisions , bacteria and archaea are known as prokaryotes , and have no membrane bound organelles within . consider the following : some diseases can be traced back to organelle lack / malformation . for example , inclusion-cell ( i-cell ) disease occurs due to a defect in the golgi . in order to mark enzymes that should be sent to lysosomes to help degrade unwanted molecules , the golgi has to bind them with a mannose 6-phosphate tag , like a shipping label . however , in patients with i-cell disease , one of the proteins that make this tag is mutated , and can not do its job , like a broken label machine . this means that proteins can not be targeted to lysosomes . these untagged proteins are the enzymes that are responsible for chopping up other proteins . what happens is the inactivated enzymes end up being sent outside the cell , while lysosomes clog up with undigested material . this disease is congenital , and usually fatal before patients reach 7 years of age . an interesting idea is that mitochondria can be used to trace maternal ancestry . since mitochondria are self-replicating and have their own dna , they are not determined by the genes found in the nucleus . instead , your mitochondria have developed from the mitochondria present in the female ovum ( egg ) that you developed from . defects in mitochondrial dna cause hereditary diseases that pass only from mother to children .
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however , despite this vast range in size , shape , and function , all these little factories have the same basic machinery . there are two main types of cells , prokaryotic and eukaryotic . prokaryotes are cells that do not have membrane bound nuclei , whereas eukaryotes do .
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why does't this article go into depth about the differences between eukaryotic cells and prokaryotic cells ?
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what is a cell right now your body is doing a million things at once . it ’ s sending electrical impulses , pumping blood , filtering urine , digesting food , making protein , storing fat , and that ’ s just the stuff you ’ re not thinking about ! you can do all this because you are made of cells — tiny units of life that are like specialized factories , full of machinery designed to accomplish the business of life . cells make up every living thing , from blue whales to the archaebacteria that live inside volcanos . just like the organisms they make up , cells can come in all shapes and sizes . nerve cells in giant squids can reach up to 12m [ 39 ft ] in length , while human eggs ( the largest human cells ) are about 0.1mm across . plant cells have protective walls made of cellulose ( which also makes up the strings in celery that make it so hard to eat ) while fungal cell walls are made from the same stuff as lobster shells . however , despite this vast range in size , shape , and function , all these little factories have the same basic machinery . there are two main types of cells , prokaryotic and eukaryotic . prokaryotes are cells that do not have membrane bound nuclei , whereas eukaryotes do . the rest of our discussion will strictly be on eukaryotes . think about what a factory needs in order to function effectively . at its most basic , a factory needs a building , a product , and a way to make that product . all cells have membranes ( the building ) , dna ( the various blueprints ) , and ribosomes ( the production line ) , and so are able to make proteins ( the product - let ’ s say we ’ re making toys ) . this article will focus on eukaryotes , since they are the cell type that contains organelles . what ’ s found inside a cell an organelle ( think of it as a cell ’ s internal organ ) is a membrane bound structure found within a cell . just like cells have membranes to hold everything in , these mini-organs are also bound in a double layer of phospholipids to insulate their little compartments within the larger cells . you can think of organelles as smaller rooms within the factory , with specialized conditions to help these rooms carry out their specific task ( like a break room stocked with goodies or a research room with cool gadgets and a special air filter ) . these organelles are found in the cytoplasm , a viscous liquid found within the cell membrane that houses the organelles and is the location of most of the action happening in a cell . below is a table of the organelles found in the basic human cell , which we ’ ll be using as our template for this discussion . organelle | function | factory part : - : | : - : | : - : nucleus | dna storage | room where the blueprints are kept mitochondrion | energy production | powerplant smooth endoplasmic reticulum ( ser ) | lipid production ; detoxification | accessory production - makes decorations for the toy , etc . rough endoplasmic reticulum ( rer ) | protein production ; in particular for export out of the cell | primary production line - makes the toys golgi apparatus | protein modification and export | shipping department peroxisome | lipid destruction ; contains oxidative enzymes | security and waste removal lysosome | protein destruction | recycling and security nucleus our dna has the blueprints for every protein in our body , all packaged into a neat double helix . the processes to transform dna into proteins are known as transcription and translation , and happen in different compartments within the cell . the first step , transcription , happens in the nucleus , which holds our dna . a membrane called the nuclear envelope surrounds the nucleus , and its job is to create a room within the cell to both protect the genetic information and to house all the molecules that are involved in processing and protecting that info . this membrane is actually a set of two lipid bilayers , so there are four sheets of lipids separating the inside of the nucleus from the cytoplasm . the space between the two bilayers is known as the perinuclear space . though part of the function of the nucleus is to separate the dna from the rest of the cell , molecules must still be able to move in and out ( e.g. , rna ) . proteins channels known as nuclear pores form holes in the nuclear envelope . the nucleus itself is filled with liquid ( called nucleoplasm ) and is similar in structure and function to cytoplasm . it is here within the nucleoplasm where chromosomes ( tightly packed strands of dna containing all our blueprints ) are found . a nucleus has interesting implications for how a cell responds to its environment . thanks to the added protection of the nuclear envelope , the dna is a little bit more secure from enzymes , pathogens , and potentially harmful products of fat and protein metabolism . since this is the only permanent copy of the instructions the cell has , it is very important to keep the dna in good condition . if the dna was not sequestered away , it would be vulnerable to damage by the aforementioned dangers , which would then lead to defective protein production . imagine a giant hole or coffee stain in the blueprint for your toy - all of a sudden you don ’ t have either enough or the right information to make a critical piece of the toy . the nuclear envelope also keeps molecules responsible for dna transcription and repair close to the dna itself - otherwise those molecules would diffuse across the entire cell and it would take a lot more work and luck to get anything done ! while transcription ( making a complementary strand of rna from dna ) is completed within the nucleus , translation ( making protein from rna instructions ) takes place in the cytoplasm . if there was no barrier between the transcription and translation machineries , it ’ s possible that poorly-made or unfinished rna would get turned into poorly made and potentially dangerous proteins . before an rna can exit the nucleus to be translated , it must get special modifications , in the form of a cap and tail at either end of the molecule , that act as a stamp of approval to let the cell know this piece of rna is complete and properly made . nucleolus within the nucleus is a small subspace known as the nucleolus . it is not bound by a membrane , so it is not an organelle . this space forms near the part of dna with instructions for making ribosomes , the molecules responsible for making proteins . ribosomes are assembled in the nucleolus , and exit the nucleus with nuclear pores . in our analogy , the robots making our product are made in a special corner of the blueprint room , before being released to the factory . endoplasmic reticulum endoplasmic means inside ( endo ) the cytoplasm ( plasm ) . reticulum comes from the latin word for net . basically , an endoplasmic reticulum is a plasma membrane found inside the cell that folds in on itself to create an internal space known as the lumen . this lumen is actually continuous with the perinuclear space , so we know the endoplasmic reticulum is attached to the nuclear envelope . there are actually two different endoplasmic reticuli in a cell : the smooth endoplasmic reticulum and the rough endoplasmic reticulum . the rough endoplasmic reticulum is the site of protein production ( where we make our major product - the toy ) while the smooth endoplasmic reticulum is where lipids ( fats ) are made ( accessories for the toy , but not the central product of the factory ) . rough endoplasmic reticulum the rough endoplasmic reticulum is so-called because its surface is studded with ribosomes , the molecules in charge of protein production . when a ribosome finds a specific rna segment , that segment may tell the ribosome to travel to the rough endoplasmic reticulum and embed itself . the protein created from this segment will find itself inside the lumen of the rough endoplasmic reticulum , where it folds and is tagged with a ( usually carbohydrate ) molecule in a process known as glycosylation that marks the protein for transport to the golgi apparatus . the rough endoplasmic reticulum is continuous with the nuclear envelope , and looks like a series of canals near the nucleus . proteins made in the rough endoplasmic reticulum as destined to either be a part of a membrane , or to be secreted from the cell membrane out of the cell . without an rough endoplasmic reticulum , it would be a lot harder to distinguish between proteins that should leave the cell , and proteins that should remain . thus , the rough endoplasmic reticulum helps cells specialize and allows for greater complexity in the organism . smooth endoplasmic reticulum the smooth endoplasmic reticulum makes lipids and steroids , instead of being involved in protein synthesis . these are fat-based molecules that are important in energy storage , membrane structure , and communication ( steroids can act as hormones ) . the smooth endoplasmic reticulum is also responsible for detoxifying the cell . it is more tubular than the rough endoplasmic reticulum , and is not necessarily continuous with the nuclear envelope . every cell has a smooth endoplasmic reticulum , but the amount will vary with cell function . for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum . golgi apparatus ( aka golgi body aka golgi ) we mentioned the golgi apparatus earlier when we discussed the production of proteins in the rough endoplasmic reticulum . if the smooth and rough endoplasmic reticula are how we make our product , the golgi is the mailroom that sends our product to customers . it is responsible for packing proteins from the rough endoplasmic reticulum into membrane-bound vesicles ( tiny compartments of lipid bilayer that store molecules ) which then translocate to the cell membrane . at the cell membrane , the vesicles can fuse with the larger lipid bilayer , causing the vesicle contents to either become part of the cell membrane or be released to the outside . different molecules actually have different fates upon entering the golgi . this determination is done by tagging the proteins with special sugar molecules that act as a shipping label for the protein . the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it . molecules in this pathway are often protein channels which allow molecules into or out of the cell , or cell identifiers which project into the extracellular space and act like a name tag for the cell . secretion : some proteins are meant to be secreted from the cell to act on other parts of the body . before these vesicles can fuse with the cell membrane , they must accumulate in number , and require a special chemical signal to be released . this way shipments only go out if they ’ re worth the cost of sending them ( you generally wouldn ’ t ship just one toy and expect to profit ) . lysosome : the final destination for proteins coming through the golgi is the lysosome . vesicles sent to this acidic organelle contain enzymes that will hydrolyze the lysosome ’ s content . lysosome the lysosome is the cell ’ s recycling center . these organelles are spheres full of enzymes ready to hydrolyze ( chop up the chemical bonds of ) whatever substance crosses the membrane , so the cell can reuse the raw material . these disposal enzymes only function properly in environments with a ph of 5 , two orders of magnitude more acidic than the cell ’ s internal ph of 7 . lysosomal proteins only being active in an acidic environment acts as safety mechanism for the rest of the cell - if the lysosome were to somehow leak or burst , the degradative enzymes would inactivate before they chopped up proteins the cell still needed . peroxisome like the lysosome , the peroxisome is a spherical organelle responsible for destroying its contents . unlike the lysosome , which mostly degrades proteins , the peroxisome is the site of fatty acid breakdown . it also protects the cell from reactive oxygen species ( ros ) molecules which could seriously damage the cell . ross are molecules like oxygen ions or peroxides that are created as a byproduct of normal cellular metabolism , but also by radiation , tobacco , and drugs . they cause what is known as oxidative stress in the cell by reacting with and damaging dna and lipid-based molecules like cell membranes . these ross are the reason we need antioxidants in our diet . mitochondria just like a factory can ’ t run without electricity , a cell can ’ t run without energy . atp ( adenosine triphosphate ) is the energy currency of the cell , and is produced in a process known as cellular respiration . though the process begins in the cytoplasm , the bulk of the energy produced comes from later steps that take place in the mitochondria . like we saw with the nuclear envelope , there are actually two lipid bilayers that separate the mitochondrial contents from the cytoplasm . we refer to them as the inner and outer mitochondrial membranes . if we cross both membranes we end up in the matrix , where pyruvate is sent after it is created from the breakdown of glucose ( this is step 1 of cellular respiration , known as glycolysis ) .the space between the two membranes is called the intermembrane space , and it has a low ph ( is acidic ) because the electron transport chain embedded in the inner membrane pumps protons ( h+ ) into it . energy to make atp comes from protons moving back into the matrix down their gradient from the intermembrane space . mitochondria are also somewhat unique in that they are self-replicating and have their own dna , almost as if they were a completely separate cell . the prevailing theory , known as the endosymbiotic theory , is that eukaryotes were first formed by large prokaryotic cells engulfing smaller cells that looked a lot like mitochondria ( and chloroplasts , more on them later ) . instead of being digested , the engulfed cells remained intact and the arrangement turned out to be advantageous to both cells , which created a symbiotic relationship . so far we ’ ve discussed organelles , the membrane-bound structures within a cell that have some sort of specialized function . now let ’ s take a moment to talk about the scaffolding that ’ s holding all of this in place - the walls and beams of our factory . cytoskeleton within the cytoplasm there is network of protein fibers known as the cytoskeleton . this structure is responsible for both cell movement and stability . the major components of the cytoskeleton are microtubules , intermediate filaments , and microfilaments . microtubules microtubules are small tubes made from the protein tubulin . these tubules are found in cilia and flagella , structures involved in cell movement . they also help provide pathways for secretory vesicles to move through the cell , and are even involved in cell division as they are a part of the mitotic spindle , which pulls homologous chromosomes apart . intermediate filaments smaller than the microtubules , but larger than the microfilaments , the intermediate filaments are made of a variety of proteins such as keratin and/or neurofilament . they are very stable , and help provide structure to the nuclear envelope and anchor organelles . microfilaments microfilaments are the thinnest part of the cytoskeleton , and are made of actin [ a highly-conserved protein that is actually the most abundant protein in most eukaryotic cells ] . actin is both flexible and strong , making it a useful protein in cell movement . in the heart , contraction is mediated through an actin-myosin system . plants and platelets so far we ’ ve covered basic organelles found in a eukaryotic cell . however , not every cell has each of these organelles , and some cells have organelles we haven ’ t discussed . for example , plant cells have chloroplasts , organelles that resemble mitochondria and are responsible for turning sunlight into useful energy for the cell ( this is like factories that are powered by energy they collect via solar panels ) . on the other hand , platelets , blood cells responsible for clotting , have no nucleus and are in fact just fragments of cytoplasm contained within a cell membrane . eukaryotes vs bacteria vs archaea it is also important to keep in mind that organelles are found only in eukaryotes , one of the three major cell divisions . the other two major divisions , bacteria and archaea are known as prokaryotes , and have no membrane bound organelles within . consider the following : some diseases can be traced back to organelle lack / malformation . for example , inclusion-cell ( i-cell ) disease occurs due to a defect in the golgi . in order to mark enzymes that should be sent to lysosomes to help degrade unwanted molecules , the golgi has to bind them with a mannose 6-phosphate tag , like a shipping label . however , in patients with i-cell disease , one of the proteins that make this tag is mutated , and can not do its job , like a broken label machine . this means that proteins can not be targeted to lysosomes . these untagged proteins are the enzymes that are responsible for chopping up other proteins . what happens is the inactivated enzymes end up being sent outside the cell , while lysosomes clog up with undigested material . this disease is congenital , and usually fatal before patients reach 7 years of age . an interesting idea is that mitochondria can be used to trace maternal ancestry . since mitochondria are self-replicating and have their own dna , they are not determined by the genes found in the nucleus . instead , your mitochondria have developed from the mitochondria present in the female ovum ( egg ) that you developed from . defects in mitochondrial dna cause hereditary diseases that pass only from mother to children .
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now let ’ s take a moment to talk about the scaffolding that ’ s holding all of this in place - the walls and beams of our factory . cytoskeleton within the cytoplasm there is network of protein fibers known as the cytoskeleton . this structure is responsible for both cell movement and stability .
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how do proteins and other items like that travel along the cytoskeleton ?
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what is a cell right now your body is doing a million things at once . it ’ s sending electrical impulses , pumping blood , filtering urine , digesting food , making protein , storing fat , and that ’ s just the stuff you ’ re not thinking about ! you can do all this because you are made of cells — tiny units of life that are like specialized factories , full of machinery designed to accomplish the business of life . cells make up every living thing , from blue whales to the archaebacteria that live inside volcanos . just like the organisms they make up , cells can come in all shapes and sizes . nerve cells in giant squids can reach up to 12m [ 39 ft ] in length , while human eggs ( the largest human cells ) are about 0.1mm across . plant cells have protective walls made of cellulose ( which also makes up the strings in celery that make it so hard to eat ) while fungal cell walls are made from the same stuff as lobster shells . however , despite this vast range in size , shape , and function , all these little factories have the same basic machinery . there are two main types of cells , prokaryotic and eukaryotic . prokaryotes are cells that do not have membrane bound nuclei , whereas eukaryotes do . the rest of our discussion will strictly be on eukaryotes . think about what a factory needs in order to function effectively . at its most basic , a factory needs a building , a product , and a way to make that product . all cells have membranes ( the building ) , dna ( the various blueprints ) , and ribosomes ( the production line ) , and so are able to make proteins ( the product - let ’ s say we ’ re making toys ) . this article will focus on eukaryotes , since they are the cell type that contains organelles . what ’ s found inside a cell an organelle ( think of it as a cell ’ s internal organ ) is a membrane bound structure found within a cell . just like cells have membranes to hold everything in , these mini-organs are also bound in a double layer of phospholipids to insulate their little compartments within the larger cells . you can think of organelles as smaller rooms within the factory , with specialized conditions to help these rooms carry out their specific task ( like a break room stocked with goodies or a research room with cool gadgets and a special air filter ) . these organelles are found in the cytoplasm , a viscous liquid found within the cell membrane that houses the organelles and is the location of most of the action happening in a cell . below is a table of the organelles found in the basic human cell , which we ’ ll be using as our template for this discussion . organelle | function | factory part : - : | : - : | : - : nucleus | dna storage | room where the blueprints are kept mitochondrion | energy production | powerplant smooth endoplasmic reticulum ( ser ) | lipid production ; detoxification | accessory production - makes decorations for the toy , etc . rough endoplasmic reticulum ( rer ) | protein production ; in particular for export out of the cell | primary production line - makes the toys golgi apparatus | protein modification and export | shipping department peroxisome | lipid destruction ; contains oxidative enzymes | security and waste removal lysosome | protein destruction | recycling and security nucleus our dna has the blueprints for every protein in our body , all packaged into a neat double helix . the processes to transform dna into proteins are known as transcription and translation , and happen in different compartments within the cell . the first step , transcription , happens in the nucleus , which holds our dna . a membrane called the nuclear envelope surrounds the nucleus , and its job is to create a room within the cell to both protect the genetic information and to house all the molecules that are involved in processing and protecting that info . this membrane is actually a set of two lipid bilayers , so there are four sheets of lipids separating the inside of the nucleus from the cytoplasm . the space between the two bilayers is known as the perinuclear space . though part of the function of the nucleus is to separate the dna from the rest of the cell , molecules must still be able to move in and out ( e.g. , rna ) . proteins channels known as nuclear pores form holes in the nuclear envelope . the nucleus itself is filled with liquid ( called nucleoplasm ) and is similar in structure and function to cytoplasm . it is here within the nucleoplasm where chromosomes ( tightly packed strands of dna containing all our blueprints ) are found . a nucleus has interesting implications for how a cell responds to its environment . thanks to the added protection of the nuclear envelope , the dna is a little bit more secure from enzymes , pathogens , and potentially harmful products of fat and protein metabolism . since this is the only permanent copy of the instructions the cell has , it is very important to keep the dna in good condition . if the dna was not sequestered away , it would be vulnerable to damage by the aforementioned dangers , which would then lead to defective protein production . imagine a giant hole or coffee stain in the blueprint for your toy - all of a sudden you don ’ t have either enough or the right information to make a critical piece of the toy . the nuclear envelope also keeps molecules responsible for dna transcription and repair close to the dna itself - otherwise those molecules would diffuse across the entire cell and it would take a lot more work and luck to get anything done ! while transcription ( making a complementary strand of rna from dna ) is completed within the nucleus , translation ( making protein from rna instructions ) takes place in the cytoplasm . if there was no barrier between the transcription and translation machineries , it ’ s possible that poorly-made or unfinished rna would get turned into poorly made and potentially dangerous proteins . before an rna can exit the nucleus to be translated , it must get special modifications , in the form of a cap and tail at either end of the molecule , that act as a stamp of approval to let the cell know this piece of rna is complete and properly made . nucleolus within the nucleus is a small subspace known as the nucleolus . it is not bound by a membrane , so it is not an organelle . this space forms near the part of dna with instructions for making ribosomes , the molecules responsible for making proteins . ribosomes are assembled in the nucleolus , and exit the nucleus with nuclear pores . in our analogy , the robots making our product are made in a special corner of the blueprint room , before being released to the factory . endoplasmic reticulum endoplasmic means inside ( endo ) the cytoplasm ( plasm ) . reticulum comes from the latin word for net . basically , an endoplasmic reticulum is a plasma membrane found inside the cell that folds in on itself to create an internal space known as the lumen . this lumen is actually continuous with the perinuclear space , so we know the endoplasmic reticulum is attached to the nuclear envelope . there are actually two different endoplasmic reticuli in a cell : the smooth endoplasmic reticulum and the rough endoplasmic reticulum . the rough endoplasmic reticulum is the site of protein production ( where we make our major product - the toy ) while the smooth endoplasmic reticulum is where lipids ( fats ) are made ( accessories for the toy , but not the central product of the factory ) . rough endoplasmic reticulum the rough endoplasmic reticulum is so-called because its surface is studded with ribosomes , the molecules in charge of protein production . when a ribosome finds a specific rna segment , that segment may tell the ribosome to travel to the rough endoplasmic reticulum and embed itself . the protein created from this segment will find itself inside the lumen of the rough endoplasmic reticulum , where it folds and is tagged with a ( usually carbohydrate ) molecule in a process known as glycosylation that marks the protein for transport to the golgi apparatus . the rough endoplasmic reticulum is continuous with the nuclear envelope , and looks like a series of canals near the nucleus . proteins made in the rough endoplasmic reticulum as destined to either be a part of a membrane , or to be secreted from the cell membrane out of the cell . without an rough endoplasmic reticulum , it would be a lot harder to distinguish between proteins that should leave the cell , and proteins that should remain . thus , the rough endoplasmic reticulum helps cells specialize and allows for greater complexity in the organism . smooth endoplasmic reticulum the smooth endoplasmic reticulum makes lipids and steroids , instead of being involved in protein synthesis . these are fat-based molecules that are important in energy storage , membrane structure , and communication ( steroids can act as hormones ) . the smooth endoplasmic reticulum is also responsible for detoxifying the cell . it is more tubular than the rough endoplasmic reticulum , and is not necessarily continuous with the nuclear envelope . every cell has a smooth endoplasmic reticulum , but the amount will vary with cell function . for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum . golgi apparatus ( aka golgi body aka golgi ) we mentioned the golgi apparatus earlier when we discussed the production of proteins in the rough endoplasmic reticulum . if the smooth and rough endoplasmic reticula are how we make our product , the golgi is the mailroom that sends our product to customers . it is responsible for packing proteins from the rough endoplasmic reticulum into membrane-bound vesicles ( tiny compartments of lipid bilayer that store molecules ) which then translocate to the cell membrane . at the cell membrane , the vesicles can fuse with the larger lipid bilayer , causing the vesicle contents to either become part of the cell membrane or be released to the outside . different molecules actually have different fates upon entering the golgi . this determination is done by tagging the proteins with special sugar molecules that act as a shipping label for the protein . the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it . molecules in this pathway are often protein channels which allow molecules into or out of the cell , or cell identifiers which project into the extracellular space and act like a name tag for the cell . secretion : some proteins are meant to be secreted from the cell to act on other parts of the body . before these vesicles can fuse with the cell membrane , they must accumulate in number , and require a special chemical signal to be released . this way shipments only go out if they ’ re worth the cost of sending them ( you generally wouldn ’ t ship just one toy and expect to profit ) . lysosome : the final destination for proteins coming through the golgi is the lysosome . vesicles sent to this acidic organelle contain enzymes that will hydrolyze the lysosome ’ s content . lysosome the lysosome is the cell ’ s recycling center . these organelles are spheres full of enzymes ready to hydrolyze ( chop up the chemical bonds of ) whatever substance crosses the membrane , so the cell can reuse the raw material . these disposal enzymes only function properly in environments with a ph of 5 , two orders of magnitude more acidic than the cell ’ s internal ph of 7 . lysosomal proteins only being active in an acidic environment acts as safety mechanism for the rest of the cell - if the lysosome were to somehow leak or burst , the degradative enzymes would inactivate before they chopped up proteins the cell still needed . peroxisome like the lysosome , the peroxisome is a spherical organelle responsible for destroying its contents . unlike the lysosome , which mostly degrades proteins , the peroxisome is the site of fatty acid breakdown . it also protects the cell from reactive oxygen species ( ros ) molecules which could seriously damage the cell . ross are molecules like oxygen ions or peroxides that are created as a byproduct of normal cellular metabolism , but also by radiation , tobacco , and drugs . they cause what is known as oxidative stress in the cell by reacting with and damaging dna and lipid-based molecules like cell membranes . these ross are the reason we need antioxidants in our diet . mitochondria just like a factory can ’ t run without electricity , a cell can ’ t run without energy . atp ( adenosine triphosphate ) is the energy currency of the cell , and is produced in a process known as cellular respiration . though the process begins in the cytoplasm , the bulk of the energy produced comes from later steps that take place in the mitochondria . like we saw with the nuclear envelope , there are actually two lipid bilayers that separate the mitochondrial contents from the cytoplasm . we refer to them as the inner and outer mitochondrial membranes . if we cross both membranes we end up in the matrix , where pyruvate is sent after it is created from the breakdown of glucose ( this is step 1 of cellular respiration , known as glycolysis ) .the space between the two membranes is called the intermembrane space , and it has a low ph ( is acidic ) because the electron transport chain embedded in the inner membrane pumps protons ( h+ ) into it . energy to make atp comes from protons moving back into the matrix down their gradient from the intermembrane space . mitochondria are also somewhat unique in that they are self-replicating and have their own dna , almost as if they were a completely separate cell . the prevailing theory , known as the endosymbiotic theory , is that eukaryotes were first formed by large prokaryotic cells engulfing smaller cells that looked a lot like mitochondria ( and chloroplasts , more on them later ) . instead of being digested , the engulfed cells remained intact and the arrangement turned out to be advantageous to both cells , which created a symbiotic relationship . so far we ’ ve discussed organelles , the membrane-bound structures within a cell that have some sort of specialized function . now let ’ s take a moment to talk about the scaffolding that ’ s holding all of this in place - the walls and beams of our factory . cytoskeleton within the cytoplasm there is network of protein fibers known as the cytoskeleton . this structure is responsible for both cell movement and stability . the major components of the cytoskeleton are microtubules , intermediate filaments , and microfilaments . microtubules microtubules are small tubes made from the protein tubulin . these tubules are found in cilia and flagella , structures involved in cell movement . they also help provide pathways for secretory vesicles to move through the cell , and are even involved in cell division as they are a part of the mitotic spindle , which pulls homologous chromosomes apart . intermediate filaments smaller than the microtubules , but larger than the microfilaments , the intermediate filaments are made of a variety of proteins such as keratin and/or neurofilament . they are very stable , and help provide structure to the nuclear envelope and anchor organelles . microfilaments microfilaments are the thinnest part of the cytoskeleton , and are made of actin [ a highly-conserved protein that is actually the most abundant protein in most eukaryotic cells ] . actin is both flexible and strong , making it a useful protein in cell movement . in the heart , contraction is mediated through an actin-myosin system . plants and platelets so far we ’ ve covered basic organelles found in a eukaryotic cell . however , not every cell has each of these organelles , and some cells have organelles we haven ’ t discussed . for example , plant cells have chloroplasts , organelles that resemble mitochondria and are responsible for turning sunlight into useful energy for the cell ( this is like factories that are powered by energy they collect via solar panels ) . on the other hand , platelets , blood cells responsible for clotting , have no nucleus and are in fact just fragments of cytoplasm contained within a cell membrane . eukaryotes vs bacteria vs archaea it is also important to keep in mind that organelles are found only in eukaryotes , one of the three major cell divisions . the other two major divisions , bacteria and archaea are known as prokaryotes , and have no membrane bound organelles within . consider the following : some diseases can be traced back to organelle lack / malformation . for example , inclusion-cell ( i-cell ) disease occurs due to a defect in the golgi . in order to mark enzymes that should be sent to lysosomes to help degrade unwanted molecules , the golgi has to bind them with a mannose 6-phosphate tag , like a shipping label . however , in patients with i-cell disease , one of the proteins that make this tag is mutated , and can not do its job , like a broken label machine . this means that proteins can not be targeted to lysosomes . these untagged proteins are the enzymes that are responsible for chopping up other proteins . what happens is the inactivated enzymes end up being sent outside the cell , while lysosomes clog up with undigested material . this disease is congenital , and usually fatal before patients reach 7 years of age . an interesting idea is that mitochondria can be used to trace maternal ancestry . since mitochondria are self-replicating and have their own dna , they are not determined by the genes found in the nucleus . instead , your mitochondria have developed from the mitochondria present in the female ovum ( egg ) that you developed from . defects in mitochondrial dna cause hereditary diseases that pass only from mother to children .
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they also help provide pathways for secretory vesicles to move through the cell , and are even involved in cell division as they are a part of the mitotic spindle , which pulls homologous chromosomes apart . intermediate filaments smaller than the microtubules , but larger than the microfilaments , the intermediate filaments are made of a variety of proteins such as keratin and/or neurofilament . they are very stable , and help provide structure to the nuclear envelope and anchor organelles .
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how are intermediate filaments providing structure if they are extremely small compared to other organelles and are just floating around ?
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what is a cell right now your body is doing a million things at once . it ’ s sending electrical impulses , pumping blood , filtering urine , digesting food , making protein , storing fat , and that ’ s just the stuff you ’ re not thinking about ! you can do all this because you are made of cells — tiny units of life that are like specialized factories , full of machinery designed to accomplish the business of life . cells make up every living thing , from blue whales to the archaebacteria that live inside volcanos . just like the organisms they make up , cells can come in all shapes and sizes . nerve cells in giant squids can reach up to 12m [ 39 ft ] in length , while human eggs ( the largest human cells ) are about 0.1mm across . plant cells have protective walls made of cellulose ( which also makes up the strings in celery that make it so hard to eat ) while fungal cell walls are made from the same stuff as lobster shells . however , despite this vast range in size , shape , and function , all these little factories have the same basic machinery . there are two main types of cells , prokaryotic and eukaryotic . prokaryotes are cells that do not have membrane bound nuclei , whereas eukaryotes do . the rest of our discussion will strictly be on eukaryotes . think about what a factory needs in order to function effectively . at its most basic , a factory needs a building , a product , and a way to make that product . all cells have membranes ( the building ) , dna ( the various blueprints ) , and ribosomes ( the production line ) , and so are able to make proteins ( the product - let ’ s say we ’ re making toys ) . this article will focus on eukaryotes , since they are the cell type that contains organelles . what ’ s found inside a cell an organelle ( think of it as a cell ’ s internal organ ) is a membrane bound structure found within a cell . just like cells have membranes to hold everything in , these mini-organs are also bound in a double layer of phospholipids to insulate their little compartments within the larger cells . you can think of organelles as smaller rooms within the factory , with specialized conditions to help these rooms carry out their specific task ( like a break room stocked with goodies or a research room with cool gadgets and a special air filter ) . these organelles are found in the cytoplasm , a viscous liquid found within the cell membrane that houses the organelles and is the location of most of the action happening in a cell . below is a table of the organelles found in the basic human cell , which we ’ ll be using as our template for this discussion . organelle | function | factory part : - : | : - : | : - : nucleus | dna storage | room where the blueprints are kept mitochondrion | energy production | powerplant smooth endoplasmic reticulum ( ser ) | lipid production ; detoxification | accessory production - makes decorations for the toy , etc . rough endoplasmic reticulum ( rer ) | protein production ; in particular for export out of the cell | primary production line - makes the toys golgi apparatus | protein modification and export | shipping department peroxisome | lipid destruction ; contains oxidative enzymes | security and waste removal lysosome | protein destruction | recycling and security nucleus our dna has the blueprints for every protein in our body , all packaged into a neat double helix . the processes to transform dna into proteins are known as transcription and translation , and happen in different compartments within the cell . the first step , transcription , happens in the nucleus , which holds our dna . a membrane called the nuclear envelope surrounds the nucleus , and its job is to create a room within the cell to both protect the genetic information and to house all the molecules that are involved in processing and protecting that info . this membrane is actually a set of two lipid bilayers , so there are four sheets of lipids separating the inside of the nucleus from the cytoplasm . the space between the two bilayers is known as the perinuclear space . though part of the function of the nucleus is to separate the dna from the rest of the cell , molecules must still be able to move in and out ( e.g. , rna ) . proteins channels known as nuclear pores form holes in the nuclear envelope . the nucleus itself is filled with liquid ( called nucleoplasm ) and is similar in structure and function to cytoplasm . it is here within the nucleoplasm where chromosomes ( tightly packed strands of dna containing all our blueprints ) are found . a nucleus has interesting implications for how a cell responds to its environment . thanks to the added protection of the nuclear envelope , the dna is a little bit more secure from enzymes , pathogens , and potentially harmful products of fat and protein metabolism . since this is the only permanent copy of the instructions the cell has , it is very important to keep the dna in good condition . if the dna was not sequestered away , it would be vulnerable to damage by the aforementioned dangers , which would then lead to defective protein production . imagine a giant hole or coffee stain in the blueprint for your toy - all of a sudden you don ’ t have either enough or the right information to make a critical piece of the toy . the nuclear envelope also keeps molecules responsible for dna transcription and repair close to the dna itself - otherwise those molecules would diffuse across the entire cell and it would take a lot more work and luck to get anything done ! while transcription ( making a complementary strand of rna from dna ) is completed within the nucleus , translation ( making protein from rna instructions ) takes place in the cytoplasm . if there was no barrier between the transcription and translation machineries , it ’ s possible that poorly-made or unfinished rna would get turned into poorly made and potentially dangerous proteins . before an rna can exit the nucleus to be translated , it must get special modifications , in the form of a cap and tail at either end of the molecule , that act as a stamp of approval to let the cell know this piece of rna is complete and properly made . nucleolus within the nucleus is a small subspace known as the nucleolus . it is not bound by a membrane , so it is not an organelle . this space forms near the part of dna with instructions for making ribosomes , the molecules responsible for making proteins . ribosomes are assembled in the nucleolus , and exit the nucleus with nuclear pores . in our analogy , the robots making our product are made in a special corner of the blueprint room , before being released to the factory . endoplasmic reticulum endoplasmic means inside ( endo ) the cytoplasm ( plasm ) . reticulum comes from the latin word for net . basically , an endoplasmic reticulum is a plasma membrane found inside the cell that folds in on itself to create an internal space known as the lumen . this lumen is actually continuous with the perinuclear space , so we know the endoplasmic reticulum is attached to the nuclear envelope . there are actually two different endoplasmic reticuli in a cell : the smooth endoplasmic reticulum and the rough endoplasmic reticulum . the rough endoplasmic reticulum is the site of protein production ( where we make our major product - the toy ) while the smooth endoplasmic reticulum is where lipids ( fats ) are made ( accessories for the toy , but not the central product of the factory ) . rough endoplasmic reticulum the rough endoplasmic reticulum is so-called because its surface is studded with ribosomes , the molecules in charge of protein production . when a ribosome finds a specific rna segment , that segment may tell the ribosome to travel to the rough endoplasmic reticulum and embed itself . the protein created from this segment will find itself inside the lumen of the rough endoplasmic reticulum , where it folds and is tagged with a ( usually carbohydrate ) molecule in a process known as glycosylation that marks the protein for transport to the golgi apparatus . the rough endoplasmic reticulum is continuous with the nuclear envelope , and looks like a series of canals near the nucleus . proteins made in the rough endoplasmic reticulum as destined to either be a part of a membrane , or to be secreted from the cell membrane out of the cell . without an rough endoplasmic reticulum , it would be a lot harder to distinguish between proteins that should leave the cell , and proteins that should remain . thus , the rough endoplasmic reticulum helps cells specialize and allows for greater complexity in the organism . smooth endoplasmic reticulum the smooth endoplasmic reticulum makes lipids and steroids , instead of being involved in protein synthesis . these are fat-based molecules that are important in energy storage , membrane structure , and communication ( steroids can act as hormones ) . the smooth endoplasmic reticulum is also responsible for detoxifying the cell . it is more tubular than the rough endoplasmic reticulum , and is not necessarily continuous with the nuclear envelope . every cell has a smooth endoplasmic reticulum , but the amount will vary with cell function . for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum . golgi apparatus ( aka golgi body aka golgi ) we mentioned the golgi apparatus earlier when we discussed the production of proteins in the rough endoplasmic reticulum . if the smooth and rough endoplasmic reticula are how we make our product , the golgi is the mailroom that sends our product to customers . it is responsible for packing proteins from the rough endoplasmic reticulum into membrane-bound vesicles ( tiny compartments of lipid bilayer that store molecules ) which then translocate to the cell membrane . at the cell membrane , the vesicles can fuse with the larger lipid bilayer , causing the vesicle contents to either become part of the cell membrane or be released to the outside . different molecules actually have different fates upon entering the golgi . this determination is done by tagging the proteins with special sugar molecules that act as a shipping label for the protein . the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it . molecules in this pathway are often protein channels which allow molecules into or out of the cell , or cell identifiers which project into the extracellular space and act like a name tag for the cell . secretion : some proteins are meant to be secreted from the cell to act on other parts of the body . before these vesicles can fuse with the cell membrane , they must accumulate in number , and require a special chemical signal to be released . this way shipments only go out if they ’ re worth the cost of sending them ( you generally wouldn ’ t ship just one toy and expect to profit ) . lysosome : the final destination for proteins coming through the golgi is the lysosome . vesicles sent to this acidic organelle contain enzymes that will hydrolyze the lysosome ’ s content . lysosome the lysosome is the cell ’ s recycling center . these organelles are spheres full of enzymes ready to hydrolyze ( chop up the chemical bonds of ) whatever substance crosses the membrane , so the cell can reuse the raw material . these disposal enzymes only function properly in environments with a ph of 5 , two orders of magnitude more acidic than the cell ’ s internal ph of 7 . lysosomal proteins only being active in an acidic environment acts as safety mechanism for the rest of the cell - if the lysosome were to somehow leak or burst , the degradative enzymes would inactivate before they chopped up proteins the cell still needed . peroxisome like the lysosome , the peroxisome is a spherical organelle responsible for destroying its contents . unlike the lysosome , which mostly degrades proteins , the peroxisome is the site of fatty acid breakdown . it also protects the cell from reactive oxygen species ( ros ) molecules which could seriously damage the cell . ross are molecules like oxygen ions or peroxides that are created as a byproduct of normal cellular metabolism , but also by radiation , tobacco , and drugs . they cause what is known as oxidative stress in the cell by reacting with and damaging dna and lipid-based molecules like cell membranes . these ross are the reason we need antioxidants in our diet . mitochondria just like a factory can ’ t run without electricity , a cell can ’ t run without energy . atp ( adenosine triphosphate ) is the energy currency of the cell , and is produced in a process known as cellular respiration . though the process begins in the cytoplasm , the bulk of the energy produced comes from later steps that take place in the mitochondria . like we saw with the nuclear envelope , there are actually two lipid bilayers that separate the mitochondrial contents from the cytoplasm . we refer to them as the inner and outer mitochondrial membranes . if we cross both membranes we end up in the matrix , where pyruvate is sent after it is created from the breakdown of glucose ( this is step 1 of cellular respiration , known as glycolysis ) .the space between the two membranes is called the intermembrane space , and it has a low ph ( is acidic ) because the electron transport chain embedded in the inner membrane pumps protons ( h+ ) into it . energy to make atp comes from protons moving back into the matrix down their gradient from the intermembrane space . mitochondria are also somewhat unique in that they are self-replicating and have their own dna , almost as if they were a completely separate cell . the prevailing theory , known as the endosymbiotic theory , is that eukaryotes were first formed by large prokaryotic cells engulfing smaller cells that looked a lot like mitochondria ( and chloroplasts , more on them later ) . instead of being digested , the engulfed cells remained intact and the arrangement turned out to be advantageous to both cells , which created a symbiotic relationship . so far we ’ ve discussed organelles , the membrane-bound structures within a cell that have some sort of specialized function . now let ’ s take a moment to talk about the scaffolding that ’ s holding all of this in place - the walls and beams of our factory . cytoskeleton within the cytoplasm there is network of protein fibers known as the cytoskeleton . this structure is responsible for both cell movement and stability . the major components of the cytoskeleton are microtubules , intermediate filaments , and microfilaments . microtubules microtubules are small tubes made from the protein tubulin . these tubules are found in cilia and flagella , structures involved in cell movement . they also help provide pathways for secretory vesicles to move through the cell , and are even involved in cell division as they are a part of the mitotic spindle , which pulls homologous chromosomes apart . intermediate filaments smaller than the microtubules , but larger than the microfilaments , the intermediate filaments are made of a variety of proteins such as keratin and/or neurofilament . they are very stable , and help provide structure to the nuclear envelope and anchor organelles . microfilaments microfilaments are the thinnest part of the cytoskeleton , and are made of actin [ a highly-conserved protein that is actually the most abundant protein in most eukaryotic cells ] . actin is both flexible and strong , making it a useful protein in cell movement . in the heart , contraction is mediated through an actin-myosin system . plants and platelets so far we ’ ve covered basic organelles found in a eukaryotic cell . however , not every cell has each of these organelles , and some cells have organelles we haven ’ t discussed . for example , plant cells have chloroplasts , organelles that resemble mitochondria and are responsible for turning sunlight into useful energy for the cell ( this is like factories that are powered by energy they collect via solar panels ) . on the other hand , platelets , blood cells responsible for clotting , have no nucleus and are in fact just fragments of cytoplasm contained within a cell membrane . eukaryotes vs bacteria vs archaea it is also important to keep in mind that organelles are found only in eukaryotes , one of the three major cell divisions . the other two major divisions , bacteria and archaea are known as prokaryotes , and have no membrane bound organelles within . consider the following : some diseases can be traced back to organelle lack / malformation . for example , inclusion-cell ( i-cell ) disease occurs due to a defect in the golgi . in order to mark enzymes that should be sent to lysosomes to help degrade unwanted molecules , the golgi has to bind them with a mannose 6-phosphate tag , like a shipping label . however , in patients with i-cell disease , one of the proteins that make this tag is mutated , and can not do its job , like a broken label machine . this means that proteins can not be targeted to lysosomes . these untagged proteins are the enzymes that are responsible for chopping up other proteins . what happens is the inactivated enzymes end up being sent outside the cell , while lysosomes clog up with undigested material . this disease is congenital , and usually fatal before patients reach 7 years of age . an interesting idea is that mitochondria can be used to trace maternal ancestry . since mitochondria are self-replicating and have their own dna , they are not determined by the genes found in the nucleus . instead , your mitochondria have developed from the mitochondria present in the female ovum ( egg ) that you developed from . defects in mitochondrial dna cause hereditary diseases that pass only from mother to children .
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the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it .
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how long does it take to go through the process of making proteins in a cell ?
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what is a cell right now your body is doing a million things at once . it ’ s sending electrical impulses , pumping blood , filtering urine , digesting food , making protein , storing fat , and that ’ s just the stuff you ’ re not thinking about ! you can do all this because you are made of cells — tiny units of life that are like specialized factories , full of machinery designed to accomplish the business of life . cells make up every living thing , from blue whales to the archaebacteria that live inside volcanos . just like the organisms they make up , cells can come in all shapes and sizes . nerve cells in giant squids can reach up to 12m [ 39 ft ] in length , while human eggs ( the largest human cells ) are about 0.1mm across . plant cells have protective walls made of cellulose ( which also makes up the strings in celery that make it so hard to eat ) while fungal cell walls are made from the same stuff as lobster shells . however , despite this vast range in size , shape , and function , all these little factories have the same basic machinery . there are two main types of cells , prokaryotic and eukaryotic . prokaryotes are cells that do not have membrane bound nuclei , whereas eukaryotes do . the rest of our discussion will strictly be on eukaryotes . think about what a factory needs in order to function effectively . at its most basic , a factory needs a building , a product , and a way to make that product . all cells have membranes ( the building ) , dna ( the various blueprints ) , and ribosomes ( the production line ) , and so are able to make proteins ( the product - let ’ s say we ’ re making toys ) . this article will focus on eukaryotes , since they are the cell type that contains organelles . what ’ s found inside a cell an organelle ( think of it as a cell ’ s internal organ ) is a membrane bound structure found within a cell . just like cells have membranes to hold everything in , these mini-organs are also bound in a double layer of phospholipids to insulate their little compartments within the larger cells . you can think of organelles as smaller rooms within the factory , with specialized conditions to help these rooms carry out their specific task ( like a break room stocked with goodies or a research room with cool gadgets and a special air filter ) . these organelles are found in the cytoplasm , a viscous liquid found within the cell membrane that houses the organelles and is the location of most of the action happening in a cell . below is a table of the organelles found in the basic human cell , which we ’ ll be using as our template for this discussion . organelle | function | factory part : - : | : - : | : - : nucleus | dna storage | room where the blueprints are kept mitochondrion | energy production | powerplant smooth endoplasmic reticulum ( ser ) | lipid production ; detoxification | accessory production - makes decorations for the toy , etc . rough endoplasmic reticulum ( rer ) | protein production ; in particular for export out of the cell | primary production line - makes the toys golgi apparatus | protein modification and export | shipping department peroxisome | lipid destruction ; contains oxidative enzymes | security and waste removal lysosome | protein destruction | recycling and security nucleus our dna has the blueprints for every protein in our body , all packaged into a neat double helix . the processes to transform dna into proteins are known as transcription and translation , and happen in different compartments within the cell . the first step , transcription , happens in the nucleus , which holds our dna . a membrane called the nuclear envelope surrounds the nucleus , and its job is to create a room within the cell to both protect the genetic information and to house all the molecules that are involved in processing and protecting that info . this membrane is actually a set of two lipid bilayers , so there are four sheets of lipids separating the inside of the nucleus from the cytoplasm . the space between the two bilayers is known as the perinuclear space . though part of the function of the nucleus is to separate the dna from the rest of the cell , molecules must still be able to move in and out ( e.g. , rna ) . proteins channels known as nuclear pores form holes in the nuclear envelope . the nucleus itself is filled with liquid ( called nucleoplasm ) and is similar in structure and function to cytoplasm . it is here within the nucleoplasm where chromosomes ( tightly packed strands of dna containing all our blueprints ) are found . a nucleus has interesting implications for how a cell responds to its environment . thanks to the added protection of the nuclear envelope , the dna is a little bit more secure from enzymes , pathogens , and potentially harmful products of fat and protein metabolism . since this is the only permanent copy of the instructions the cell has , it is very important to keep the dna in good condition . if the dna was not sequestered away , it would be vulnerable to damage by the aforementioned dangers , which would then lead to defective protein production . imagine a giant hole or coffee stain in the blueprint for your toy - all of a sudden you don ’ t have either enough or the right information to make a critical piece of the toy . the nuclear envelope also keeps molecules responsible for dna transcription and repair close to the dna itself - otherwise those molecules would diffuse across the entire cell and it would take a lot more work and luck to get anything done ! while transcription ( making a complementary strand of rna from dna ) is completed within the nucleus , translation ( making protein from rna instructions ) takes place in the cytoplasm . if there was no barrier between the transcription and translation machineries , it ’ s possible that poorly-made or unfinished rna would get turned into poorly made and potentially dangerous proteins . before an rna can exit the nucleus to be translated , it must get special modifications , in the form of a cap and tail at either end of the molecule , that act as a stamp of approval to let the cell know this piece of rna is complete and properly made . nucleolus within the nucleus is a small subspace known as the nucleolus . it is not bound by a membrane , so it is not an organelle . this space forms near the part of dna with instructions for making ribosomes , the molecules responsible for making proteins . ribosomes are assembled in the nucleolus , and exit the nucleus with nuclear pores . in our analogy , the robots making our product are made in a special corner of the blueprint room , before being released to the factory . endoplasmic reticulum endoplasmic means inside ( endo ) the cytoplasm ( plasm ) . reticulum comes from the latin word for net . basically , an endoplasmic reticulum is a plasma membrane found inside the cell that folds in on itself to create an internal space known as the lumen . this lumen is actually continuous with the perinuclear space , so we know the endoplasmic reticulum is attached to the nuclear envelope . there are actually two different endoplasmic reticuli in a cell : the smooth endoplasmic reticulum and the rough endoplasmic reticulum . the rough endoplasmic reticulum is the site of protein production ( where we make our major product - the toy ) while the smooth endoplasmic reticulum is where lipids ( fats ) are made ( accessories for the toy , but not the central product of the factory ) . rough endoplasmic reticulum the rough endoplasmic reticulum is so-called because its surface is studded with ribosomes , the molecules in charge of protein production . when a ribosome finds a specific rna segment , that segment may tell the ribosome to travel to the rough endoplasmic reticulum and embed itself . the protein created from this segment will find itself inside the lumen of the rough endoplasmic reticulum , where it folds and is tagged with a ( usually carbohydrate ) molecule in a process known as glycosylation that marks the protein for transport to the golgi apparatus . the rough endoplasmic reticulum is continuous with the nuclear envelope , and looks like a series of canals near the nucleus . proteins made in the rough endoplasmic reticulum as destined to either be a part of a membrane , or to be secreted from the cell membrane out of the cell . without an rough endoplasmic reticulum , it would be a lot harder to distinguish between proteins that should leave the cell , and proteins that should remain . thus , the rough endoplasmic reticulum helps cells specialize and allows for greater complexity in the organism . smooth endoplasmic reticulum the smooth endoplasmic reticulum makes lipids and steroids , instead of being involved in protein synthesis . these are fat-based molecules that are important in energy storage , membrane structure , and communication ( steroids can act as hormones ) . the smooth endoplasmic reticulum is also responsible for detoxifying the cell . it is more tubular than the rough endoplasmic reticulum , and is not necessarily continuous with the nuclear envelope . every cell has a smooth endoplasmic reticulum , but the amount will vary with cell function . for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum . golgi apparatus ( aka golgi body aka golgi ) we mentioned the golgi apparatus earlier when we discussed the production of proteins in the rough endoplasmic reticulum . if the smooth and rough endoplasmic reticula are how we make our product , the golgi is the mailroom that sends our product to customers . it is responsible for packing proteins from the rough endoplasmic reticulum into membrane-bound vesicles ( tiny compartments of lipid bilayer that store molecules ) which then translocate to the cell membrane . at the cell membrane , the vesicles can fuse with the larger lipid bilayer , causing the vesicle contents to either become part of the cell membrane or be released to the outside . different molecules actually have different fates upon entering the golgi . this determination is done by tagging the proteins with special sugar molecules that act as a shipping label for the protein . the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it . molecules in this pathway are often protein channels which allow molecules into or out of the cell , or cell identifiers which project into the extracellular space and act like a name tag for the cell . secretion : some proteins are meant to be secreted from the cell to act on other parts of the body . before these vesicles can fuse with the cell membrane , they must accumulate in number , and require a special chemical signal to be released . this way shipments only go out if they ’ re worth the cost of sending them ( you generally wouldn ’ t ship just one toy and expect to profit ) . lysosome : the final destination for proteins coming through the golgi is the lysosome . vesicles sent to this acidic organelle contain enzymes that will hydrolyze the lysosome ’ s content . lysosome the lysosome is the cell ’ s recycling center . these organelles are spheres full of enzymes ready to hydrolyze ( chop up the chemical bonds of ) whatever substance crosses the membrane , so the cell can reuse the raw material . these disposal enzymes only function properly in environments with a ph of 5 , two orders of magnitude more acidic than the cell ’ s internal ph of 7 . lysosomal proteins only being active in an acidic environment acts as safety mechanism for the rest of the cell - if the lysosome were to somehow leak or burst , the degradative enzymes would inactivate before they chopped up proteins the cell still needed . peroxisome like the lysosome , the peroxisome is a spherical organelle responsible for destroying its contents . unlike the lysosome , which mostly degrades proteins , the peroxisome is the site of fatty acid breakdown . it also protects the cell from reactive oxygen species ( ros ) molecules which could seriously damage the cell . ross are molecules like oxygen ions or peroxides that are created as a byproduct of normal cellular metabolism , but also by radiation , tobacco , and drugs . they cause what is known as oxidative stress in the cell by reacting with and damaging dna and lipid-based molecules like cell membranes . these ross are the reason we need antioxidants in our diet . mitochondria just like a factory can ’ t run without electricity , a cell can ’ t run without energy . atp ( adenosine triphosphate ) is the energy currency of the cell , and is produced in a process known as cellular respiration . though the process begins in the cytoplasm , the bulk of the energy produced comes from later steps that take place in the mitochondria . like we saw with the nuclear envelope , there are actually two lipid bilayers that separate the mitochondrial contents from the cytoplasm . we refer to them as the inner and outer mitochondrial membranes . if we cross both membranes we end up in the matrix , where pyruvate is sent after it is created from the breakdown of glucose ( this is step 1 of cellular respiration , known as glycolysis ) .the space between the two membranes is called the intermembrane space , and it has a low ph ( is acidic ) because the electron transport chain embedded in the inner membrane pumps protons ( h+ ) into it . energy to make atp comes from protons moving back into the matrix down their gradient from the intermembrane space . mitochondria are also somewhat unique in that they are self-replicating and have their own dna , almost as if they were a completely separate cell . the prevailing theory , known as the endosymbiotic theory , is that eukaryotes were first formed by large prokaryotic cells engulfing smaller cells that looked a lot like mitochondria ( and chloroplasts , more on them later ) . instead of being digested , the engulfed cells remained intact and the arrangement turned out to be advantageous to both cells , which created a symbiotic relationship . so far we ’ ve discussed organelles , the membrane-bound structures within a cell that have some sort of specialized function . now let ’ s take a moment to talk about the scaffolding that ’ s holding all of this in place - the walls and beams of our factory . cytoskeleton within the cytoplasm there is network of protein fibers known as the cytoskeleton . this structure is responsible for both cell movement and stability . the major components of the cytoskeleton are microtubules , intermediate filaments , and microfilaments . microtubules microtubules are small tubes made from the protein tubulin . these tubules are found in cilia and flagella , structures involved in cell movement . they also help provide pathways for secretory vesicles to move through the cell , and are even involved in cell division as they are a part of the mitotic spindle , which pulls homologous chromosomes apart . intermediate filaments smaller than the microtubules , but larger than the microfilaments , the intermediate filaments are made of a variety of proteins such as keratin and/or neurofilament . they are very stable , and help provide structure to the nuclear envelope and anchor organelles . microfilaments microfilaments are the thinnest part of the cytoskeleton , and are made of actin [ a highly-conserved protein that is actually the most abundant protein in most eukaryotic cells ] . actin is both flexible and strong , making it a useful protein in cell movement . in the heart , contraction is mediated through an actin-myosin system . plants and platelets so far we ’ ve covered basic organelles found in a eukaryotic cell . however , not every cell has each of these organelles , and some cells have organelles we haven ’ t discussed . for example , plant cells have chloroplasts , organelles that resemble mitochondria and are responsible for turning sunlight into useful energy for the cell ( this is like factories that are powered by energy they collect via solar panels ) . on the other hand , platelets , blood cells responsible for clotting , have no nucleus and are in fact just fragments of cytoplasm contained within a cell membrane . eukaryotes vs bacteria vs archaea it is also important to keep in mind that organelles are found only in eukaryotes , one of the three major cell divisions . the other two major divisions , bacteria and archaea are known as prokaryotes , and have no membrane bound organelles within . consider the following : some diseases can be traced back to organelle lack / malformation . for example , inclusion-cell ( i-cell ) disease occurs due to a defect in the golgi . in order to mark enzymes that should be sent to lysosomes to help degrade unwanted molecules , the golgi has to bind them with a mannose 6-phosphate tag , like a shipping label . however , in patients with i-cell disease , one of the proteins that make this tag is mutated , and can not do its job , like a broken label machine . this means that proteins can not be targeted to lysosomes . these untagged proteins are the enzymes that are responsible for chopping up other proteins . what happens is the inactivated enzymes end up being sent outside the cell , while lysosomes clog up with undigested material . this disease is congenital , and usually fatal before patients reach 7 years of age . an interesting idea is that mitochondria can be used to trace maternal ancestry . since mitochondria are self-replicating and have their own dna , they are not determined by the genes found in the nucleus . instead , your mitochondria have developed from the mitochondria present in the female ovum ( egg ) that you developed from . defects in mitochondrial dna cause hereditary diseases that pass only from mother to children .
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plants and platelets so far we ’ ve covered basic organelles found in a eukaryotic cell . however , not every cell has each of these organelles , and some cells have organelles we haven ’ t discussed . for example , plant cells have chloroplasts , organelles that resemble mitochondria and are responsible for turning sunlight into useful energy for the cell ( this is like factories that are powered by energy they collect via solar panels ) .
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why do some cells in the same plant end up with different sets of organelles ?
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what is a cell right now your body is doing a million things at once . it ’ s sending electrical impulses , pumping blood , filtering urine , digesting food , making protein , storing fat , and that ’ s just the stuff you ’ re not thinking about ! you can do all this because you are made of cells — tiny units of life that are like specialized factories , full of machinery designed to accomplish the business of life . cells make up every living thing , from blue whales to the archaebacteria that live inside volcanos . just like the organisms they make up , cells can come in all shapes and sizes . nerve cells in giant squids can reach up to 12m [ 39 ft ] in length , while human eggs ( the largest human cells ) are about 0.1mm across . plant cells have protective walls made of cellulose ( which also makes up the strings in celery that make it so hard to eat ) while fungal cell walls are made from the same stuff as lobster shells . however , despite this vast range in size , shape , and function , all these little factories have the same basic machinery . there are two main types of cells , prokaryotic and eukaryotic . prokaryotes are cells that do not have membrane bound nuclei , whereas eukaryotes do . the rest of our discussion will strictly be on eukaryotes . think about what a factory needs in order to function effectively . at its most basic , a factory needs a building , a product , and a way to make that product . all cells have membranes ( the building ) , dna ( the various blueprints ) , and ribosomes ( the production line ) , and so are able to make proteins ( the product - let ’ s say we ’ re making toys ) . this article will focus on eukaryotes , since they are the cell type that contains organelles . what ’ s found inside a cell an organelle ( think of it as a cell ’ s internal organ ) is a membrane bound structure found within a cell . just like cells have membranes to hold everything in , these mini-organs are also bound in a double layer of phospholipids to insulate their little compartments within the larger cells . you can think of organelles as smaller rooms within the factory , with specialized conditions to help these rooms carry out their specific task ( like a break room stocked with goodies or a research room with cool gadgets and a special air filter ) . these organelles are found in the cytoplasm , a viscous liquid found within the cell membrane that houses the organelles and is the location of most of the action happening in a cell . below is a table of the organelles found in the basic human cell , which we ’ ll be using as our template for this discussion . organelle | function | factory part : - : | : - : | : - : nucleus | dna storage | room where the blueprints are kept mitochondrion | energy production | powerplant smooth endoplasmic reticulum ( ser ) | lipid production ; detoxification | accessory production - makes decorations for the toy , etc . rough endoplasmic reticulum ( rer ) | protein production ; in particular for export out of the cell | primary production line - makes the toys golgi apparatus | protein modification and export | shipping department peroxisome | lipid destruction ; contains oxidative enzymes | security and waste removal lysosome | protein destruction | recycling and security nucleus our dna has the blueprints for every protein in our body , all packaged into a neat double helix . the processes to transform dna into proteins are known as transcription and translation , and happen in different compartments within the cell . the first step , transcription , happens in the nucleus , which holds our dna . a membrane called the nuclear envelope surrounds the nucleus , and its job is to create a room within the cell to both protect the genetic information and to house all the molecules that are involved in processing and protecting that info . this membrane is actually a set of two lipid bilayers , so there are four sheets of lipids separating the inside of the nucleus from the cytoplasm . the space between the two bilayers is known as the perinuclear space . though part of the function of the nucleus is to separate the dna from the rest of the cell , molecules must still be able to move in and out ( e.g. , rna ) . proteins channels known as nuclear pores form holes in the nuclear envelope . the nucleus itself is filled with liquid ( called nucleoplasm ) and is similar in structure and function to cytoplasm . it is here within the nucleoplasm where chromosomes ( tightly packed strands of dna containing all our blueprints ) are found . a nucleus has interesting implications for how a cell responds to its environment . thanks to the added protection of the nuclear envelope , the dna is a little bit more secure from enzymes , pathogens , and potentially harmful products of fat and protein metabolism . since this is the only permanent copy of the instructions the cell has , it is very important to keep the dna in good condition . if the dna was not sequestered away , it would be vulnerable to damage by the aforementioned dangers , which would then lead to defective protein production . imagine a giant hole or coffee stain in the blueprint for your toy - all of a sudden you don ’ t have either enough or the right information to make a critical piece of the toy . the nuclear envelope also keeps molecules responsible for dna transcription and repair close to the dna itself - otherwise those molecules would diffuse across the entire cell and it would take a lot more work and luck to get anything done ! while transcription ( making a complementary strand of rna from dna ) is completed within the nucleus , translation ( making protein from rna instructions ) takes place in the cytoplasm . if there was no barrier between the transcription and translation machineries , it ’ s possible that poorly-made or unfinished rna would get turned into poorly made and potentially dangerous proteins . before an rna can exit the nucleus to be translated , it must get special modifications , in the form of a cap and tail at either end of the molecule , that act as a stamp of approval to let the cell know this piece of rna is complete and properly made . nucleolus within the nucleus is a small subspace known as the nucleolus . it is not bound by a membrane , so it is not an organelle . this space forms near the part of dna with instructions for making ribosomes , the molecules responsible for making proteins . ribosomes are assembled in the nucleolus , and exit the nucleus with nuclear pores . in our analogy , the robots making our product are made in a special corner of the blueprint room , before being released to the factory . endoplasmic reticulum endoplasmic means inside ( endo ) the cytoplasm ( plasm ) . reticulum comes from the latin word for net . basically , an endoplasmic reticulum is a plasma membrane found inside the cell that folds in on itself to create an internal space known as the lumen . this lumen is actually continuous with the perinuclear space , so we know the endoplasmic reticulum is attached to the nuclear envelope . there are actually two different endoplasmic reticuli in a cell : the smooth endoplasmic reticulum and the rough endoplasmic reticulum . the rough endoplasmic reticulum is the site of protein production ( where we make our major product - the toy ) while the smooth endoplasmic reticulum is where lipids ( fats ) are made ( accessories for the toy , but not the central product of the factory ) . rough endoplasmic reticulum the rough endoplasmic reticulum is so-called because its surface is studded with ribosomes , the molecules in charge of protein production . when a ribosome finds a specific rna segment , that segment may tell the ribosome to travel to the rough endoplasmic reticulum and embed itself . the protein created from this segment will find itself inside the lumen of the rough endoplasmic reticulum , where it folds and is tagged with a ( usually carbohydrate ) molecule in a process known as glycosylation that marks the protein for transport to the golgi apparatus . the rough endoplasmic reticulum is continuous with the nuclear envelope , and looks like a series of canals near the nucleus . proteins made in the rough endoplasmic reticulum as destined to either be a part of a membrane , or to be secreted from the cell membrane out of the cell . without an rough endoplasmic reticulum , it would be a lot harder to distinguish between proteins that should leave the cell , and proteins that should remain . thus , the rough endoplasmic reticulum helps cells specialize and allows for greater complexity in the organism . smooth endoplasmic reticulum the smooth endoplasmic reticulum makes lipids and steroids , instead of being involved in protein synthesis . these are fat-based molecules that are important in energy storage , membrane structure , and communication ( steroids can act as hormones ) . the smooth endoplasmic reticulum is also responsible for detoxifying the cell . it is more tubular than the rough endoplasmic reticulum , and is not necessarily continuous with the nuclear envelope . every cell has a smooth endoplasmic reticulum , but the amount will vary with cell function . for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum . golgi apparatus ( aka golgi body aka golgi ) we mentioned the golgi apparatus earlier when we discussed the production of proteins in the rough endoplasmic reticulum . if the smooth and rough endoplasmic reticula are how we make our product , the golgi is the mailroom that sends our product to customers . it is responsible for packing proteins from the rough endoplasmic reticulum into membrane-bound vesicles ( tiny compartments of lipid bilayer that store molecules ) which then translocate to the cell membrane . at the cell membrane , the vesicles can fuse with the larger lipid bilayer , causing the vesicle contents to either become part of the cell membrane or be released to the outside . different molecules actually have different fates upon entering the golgi . this determination is done by tagging the proteins with special sugar molecules that act as a shipping label for the protein . the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it . molecules in this pathway are often protein channels which allow molecules into or out of the cell , or cell identifiers which project into the extracellular space and act like a name tag for the cell . secretion : some proteins are meant to be secreted from the cell to act on other parts of the body . before these vesicles can fuse with the cell membrane , they must accumulate in number , and require a special chemical signal to be released . this way shipments only go out if they ’ re worth the cost of sending them ( you generally wouldn ’ t ship just one toy and expect to profit ) . lysosome : the final destination for proteins coming through the golgi is the lysosome . vesicles sent to this acidic organelle contain enzymes that will hydrolyze the lysosome ’ s content . lysosome the lysosome is the cell ’ s recycling center . these organelles are spheres full of enzymes ready to hydrolyze ( chop up the chemical bonds of ) whatever substance crosses the membrane , so the cell can reuse the raw material . these disposal enzymes only function properly in environments with a ph of 5 , two orders of magnitude more acidic than the cell ’ s internal ph of 7 . lysosomal proteins only being active in an acidic environment acts as safety mechanism for the rest of the cell - if the lysosome were to somehow leak or burst , the degradative enzymes would inactivate before they chopped up proteins the cell still needed . peroxisome like the lysosome , the peroxisome is a spherical organelle responsible for destroying its contents . unlike the lysosome , which mostly degrades proteins , the peroxisome is the site of fatty acid breakdown . it also protects the cell from reactive oxygen species ( ros ) molecules which could seriously damage the cell . ross are molecules like oxygen ions or peroxides that are created as a byproduct of normal cellular metabolism , but also by radiation , tobacco , and drugs . they cause what is known as oxidative stress in the cell by reacting with and damaging dna and lipid-based molecules like cell membranes . these ross are the reason we need antioxidants in our diet . mitochondria just like a factory can ’ t run without electricity , a cell can ’ t run without energy . atp ( adenosine triphosphate ) is the energy currency of the cell , and is produced in a process known as cellular respiration . though the process begins in the cytoplasm , the bulk of the energy produced comes from later steps that take place in the mitochondria . like we saw with the nuclear envelope , there are actually two lipid bilayers that separate the mitochondrial contents from the cytoplasm . we refer to them as the inner and outer mitochondrial membranes . if we cross both membranes we end up in the matrix , where pyruvate is sent after it is created from the breakdown of glucose ( this is step 1 of cellular respiration , known as glycolysis ) .the space between the two membranes is called the intermembrane space , and it has a low ph ( is acidic ) because the electron transport chain embedded in the inner membrane pumps protons ( h+ ) into it . energy to make atp comes from protons moving back into the matrix down their gradient from the intermembrane space . mitochondria are also somewhat unique in that they are self-replicating and have their own dna , almost as if they were a completely separate cell . the prevailing theory , known as the endosymbiotic theory , is that eukaryotes were first formed by large prokaryotic cells engulfing smaller cells that looked a lot like mitochondria ( and chloroplasts , more on them later ) . instead of being digested , the engulfed cells remained intact and the arrangement turned out to be advantageous to both cells , which created a symbiotic relationship . so far we ’ ve discussed organelles , the membrane-bound structures within a cell that have some sort of specialized function . now let ’ s take a moment to talk about the scaffolding that ’ s holding all of this in place - the walls and beams of our factory . cytoskeleton within the cytoplasm there is network of protein fibers known as the cytoskeleton . this structure is responsible for both cell movement and stability . the major components of the cytoskeleton are microtubules , intermediate filaments , and microfilaments . microtubules microtubules are small tubes made from the protein tubulin . these tubules are found in cilia and flagella , structures involved in cell movement . they also help provide pathways for secretory vesicles to move through the cell , and are even involved in cell division as they are a part of the mitotic spindle , which pulls homologous chromosomes apart . intermediate filaments smaller than the microtubules , but larger than the microfilaments , the intermediate filaments are made of a variety of proteins such as keratin and/or neurofilament . they are very stable , and help provide structure to the nuclear envelope and anchor organelles . microfilaments microfilaments are the thinnest part of the cytoskeleton , and are made of actin [ a highly-conserved protein that is actually the most abundant protein in most eukaryotic cells ] . actin is both flexible and strong , making it a useful protein in cell movement . in the heart , contraction is mediated through an actin-myosin system . plants and platelets so far we ’ ve covered basic organelles found in a eukaryotic cell . however , not every cell has each of these organelles , and some cells have organelles we haven ’ t discussed . for example , plant cells have chloroplasts , organelles that resemble mitochondria and are responsible for turning sunlight into useful energy for the cell ( this is like factories that are powered by energy they collect via solar panels ) . on the other hand , platelets , blood cells responsible for clotting , have no nucleus and are in fact just fragments of cytoplasm contained within a cell membrane . eukaryotes vs bacteria vs archaea it is also important to keep in mind that organelles are found only in eukaryotes , one of the three major cell divisions . the other two major divisions , bacteria and archaea are known as prokaryotes , and have no membrane bound organelles within . consider the following : some diseases can be traced back to organelle lack / malformation . for example , inclusion-cell ( i-cell ) disease occurs due to a defect in the golgi . in order to mark enzymes that should be sent to lysosomes to help degrade unwanted molecules , the golgi has to bind them with a mannose 6-phosphate tag , like a shipping label . however , in patients with i-cell disease , one of the proteins that make this tag is mutated , and can not do its job , like a broken label machine . this means that proteins can not be targeted to lysosomes . these untagged proteins are the enzymes that are responsible for chopping up other proteins . what happens is the inactivated enzymes end up being sent outside the cell , while lysosomes clog up with undigested material . this disease is congenital , and usually fatal before patients reach 7 years of age . an interesting idea is that mitochondria can be used to trace maternal ancestry . since mitochondria are self-replicating and have their own dna , they are not determined by the genes found in the nucleus . instead , your mitochondria have developed from the mitochondria present in the female ovum ( egg ) that you developed from . defects in mitochondrial dna cause hereditary diseases that pass only from mother to children .
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vesicles sent to this acidic organelle contain enzymes that will hydrolyze the lysosome ’ s content . lysosome the lysosome is the cell ’ s recycling center . these organelles are spheres full of enzymes ready to hydrolyze ( chop up the chemical bonds of ) whatever substance crosses the membrane , so the cell can reuse the raw material .
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how does the cell re-use the protein after the lysosome destroys the substance ?
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what is a cell right now your body is doing a million things at once . it ’ s sending electrical impulses , pumping blood , filtering urine , digesting food , making protein , storing fat , and that ’ s just the stuff you ’ re not thinking about ! you can do all this because you are made of cells — tiny units of life that are like specialized factories , full of machinery designed to accomplish the business of life . cells make up every living thing , from blue whales to the archaebacteria that live inside volcanos . just like the organisms they make up , cells can come in all shapes and sizes . nerve cells in giant squids can reach up to 12m [ 39 ft ] in length , while human eggs ( the largest human cells ) are about 0.1mm across . plant cells have protective walls made of cellulose ( which also makes up the strings in celery that make it so hard to eat ) while fungal cell walls are made from the same stuff as lobster shells . however , despite this vast range in size , shape , and function , all these little factories have the same basic machinery . there are two main types of cells , prokaryotic and eukaryotic . prokaryotes are cells that do not have membrane bound nuclei , whereas eukaryotes do . the rest of our discussion will strictly be on eukaryotes . think about what a factory needs in order to function effectively . at its most basic , a factory needs a building , a product , and a way to make that product . all cells have membranes ( the building ) , dna ( the various blueprints ) , and ribosomes ( the production line ) , and so are able to make proteins ( the product - let ’ s say we ’ re making toys ) . this article will focus on eukaryotes , since they are the cell type that contains organelles . what ’ s found inside a cell an organelle ( think of it as a cell ’ s internal organ ) is a membrane bound structure found within a cell . just like cells have membranes to hold everything in , these mini-organs are also bound in a double layer of phospholipids to insulate their little compartments within the larger cells . you can think of organelles as smaller rooms within the factory , with specialized conditions to help these rooms carry out their specific task ( like a break room stocked with goodies or a research room with cool gadgets and a special air filter ) . these organelles are found in the cytoplasm , a viscous liquid found within the cell membrane that houses the organelles and is the location of most of the action happening in a cell . below is a table of the organelles found in the basic human cell , which we ’ ll be using as our template for this discussion . organelle | function | factory part : - : | : - : | : - : nucleus | dna storage | room where the blueprints are kept mitochondrion | energy production | powerplant smooth endoplasmic reticulum ( ser ) | lipid production ; detoxification | accessory production - makes decorations for the toy , etc . rough endoplasmic reticulum ( rer ) | protein production ; in particular for export out of the cell | primary production line - makes the toys golgi apparatus | protein modification and export | shipping department peroxisome | lipid destruction ; contains oxidative enzymes | security and waste removal lysosome | protein destruction | recycling and security nucleus our dna has the blueprints for every protein in our body , all packaged into a neat double helix . the processes to transform dna into proteins are known as transcription and translation , and happen in different compartments within the cell . the first step , transcription , happens in the nucleus , which holds our dna . a membrane called the nuclear envelope surrounds the nucleus , and its job is to create a room within the cell to both protect the genetic information and to house all the molecules that are involved in processing and protecting that info . this membrane is actually a set of two lipid bilayers , so there are four sheets of lipids separating the inside of the nucleus from the cytoplasm . the space between the two bilayers is known as the perinuclear space . though part of the function of the nucleus is to separate the dna from the rest of the cell , molecules must still be able to move in and out ( e.g. , rna ) . proteins channels known as nuclear pores form holes in the nuclear envelope . the nucleus itself is filled with liquid ( called nucleoplasm ) and is similar in structure and function to cytoplasm . it is here within the nucleoplasm where chromosomes ( tightly packed strands of dna containing all our blueprints ) are found . a nucleus has interesting implications for how a cell responds to its environment . thanks to the added protection of the nuclear envelope , the dna is a little bit more secure from enzymes , pathogens , and potentially harmful products of fat and protein metabolism . since this is the only permanent copy of the instructions the cell has , it is very important to keep the dna in good condition . if the dna was not sequestered away , it would be vulnerable to damage by the aforementioned dangers , which would then lead to defective protein production . imagine a giant hole or coffee stain in the blueprint for your toy - all of a sudden you don ’ t have either enough or the right information to make a critical piece of the toy . the nuclear envelope also keeps molecules responsible for dna transcription and repair close to the dna itself - otherwise those molecules would diffuse across the entire cell and it would take a lot more work and luck to get anything done ! while transcription ( making a complementary strand of rna from dna ) is completed within the nucleus , translation ( making protein from rna instructions ) takes place in the cytoplasm . if there was no barrier between the transcription and translation machineries , it ’ s possible that poorly-made or unfinished rna would get turned into poorly made and potentially dangerous proteins . before an rna can exit the nucleus to be translated , it must get special modifications , in the form of a cap and tail at either end of the molecule , that act as a stamp of approval to let the cell know this piece of rna is complete and properly made . nucleolus within the nucleus is a small subspace known as the nucleolus . it is not bound by a membrane , so it is not an organelle . this space forms near the part of dna with instructions for making ribosomes , the molecules responsible for making proteins . ribosomes are assembled in the nucleolus , and exit the nucleus with nuclear pores . in our analogy , the robots making our product are made in a special corner of the blueprint room , before being released to the factory . endoplasmic reticulum endoplasmic means inside ( endo ) the cytoplasm ( plasm ) . reticulum comes from the latin word for net . basically , an endoplasmic reticulum is a plasma membrane found inside the cell that folds in on itself to create an internal space known as the lumen . this lumen is actually continuous with the perinuclear space , so we know the endoplasmic reticulum is attached to the nuclear envelope . there are actually two different endoplasmic reticuli in a cell : the smooth endoplasmic reticulum and the rough endoplasmic reticulum . the rough endoplasmic reticulum is the site of protein production ( where we make our major product - the toy ) while the smooth endoplasmic reticulum is where lipids ( fats ) are made ( accessories for the toy , but not the central product of the factory ) . rough endoplasmic reticulum the rough endoplasmic reticulum is so-called because its surface is studded with ribosomes , the molecules in charge of protein production . when a ribosome finds a specific rna segment , that segment may tell the ribosome to travel to the rough endoplasmic reticulum and embed itself . the protein created from this segment will find itself inside the lumen of the rough endoplasmic reticulum , where it folds and is tagged with a ( usually carbohydrate ) molecule in a process known as glycosylation that marks the protein for transport to the golgi apparatus . the rough endoplasmic reticulum is continuous with the nuclear envelope , and looks like a series of canals near the nucleus . proteins made in the rough endoplasmic reticulum as destined to either be a part of a membrane , or to be secreted from the cell membrane out of the cell . without an rough endoplasmic reticulum , it would be a lot harder to distinguish between proteins that should leave the cell , and proteins that should remain . thus , the rough endoplasmic reticulum helps cells specialize and allows for greater complexity in the organism . smooth endoplasmic reticulum the smooth endoplasmic reticulum makes lipids and steroids , instead of being involved in protein synthesis . these are fat-based molecules that are important in energy storage , membrane structure , and communication ( steroids can act as hormones ) . the smooth endoplasmic reticulum is also responsible for detoxifying the cell . it is more tubular than the rough endoplasmic reticulum , and is not necessarily continuous with the nuclear envelope . every cell has a smooth endoplasmic reticulum , but the amount will vary with cell function . for example , the liver , which is responsible for most of the body ’ s detoxification , has a larger amount of smooth endoplasmic reticulum . golgi apparatus ( aka golgi body aka golgi ) we mentioned the golgi apparatus earlier when we discussed the production of proteins in the rough endoplasmic reticulum . if the smooth and rough endoplasmic reticula are how we make our product , the golgi is the mailroom that sends our product to customers . it is responsible for packing proteins from the rough endoplasmic reticulum into membrane-bound vesicles ( tiny compartments of lipid bilayer that store molecules ) which then translocate to the cell membrane . at the cell membrane , the vesicles can fuse with the larger lipid bilayer , causing the vesicle contents to either become part of the cell membrane or be released to the outside . different molecules actually have different fates upon entering the golgi . this determination is done by tagging the proteins with special sugar molecules that act as a shipping label for the protein . the shipping department identifies the molecule and sets it on one of 4 paths : cytosol : the proteins that enter the golgi by mistake are sent back into the cytosol ( imagine the barcode scanning wrong and the item being returned ) . cell membrane : proteins destined for the cell membrane are processed continuously . once the vesicle is made , it moves to the cell membrane and fuses with it . molecules in this pathway are often protein channels which allow molecules into or out of the cell , or cell identifiers which project into the extracellular space and act like a name tag for the cell . secretion : some proteins are meant to be secreted from the cell to act on other parts of the body . before these vesicles can fuse with the cell membrane , they must accumulate in number , and require a special chemical signal to be released . this way shipments only go out if they ’ re worth the cost of sending them ( you generally wouldn ’ t ship just one toy and expect to profit ) . lysosome : the final destination for proteins coming through the golgi is the lysosome . vesicles sent to this acidic organelle contain enzymes that will hydrolyze the lysosome ’ s content . lysosome the lysosome is the cell ’ s recycling center . these organelles are spheres full of enzymes ready to hydrolyze ( chop up the chemical bonds of ) whatever substance crosses the membrane , so the cell can reuse the raw material . these disposal enzymes only function properly in environments with a ph of 5 , two orders of magnitude more acidic than the cell ’ s internal ph of 7 . lysosomal proteins only being active in an acidic environment acts as safety mechanism for the rest of the cell - if the lysosome were to somehow leak or burst , the degradative enzymes would inactivate before they chopped up proteins the cell still needed . peroxisome like the lysosome , the peroxisome is a spherical organelle responsible for destroying its contents . unlike the lysosome , which mostly degrades proteins , the peroxisome is the site of fatty acid breakdown . it also protects the cell from reactive oxygen species ( ros ) molecules which could seriously damage the cell . ross are molecules like oxygen ions or peroxides that are created as a byproduct of normal cellular metabolism , but also by radiation , tobacco , and drugs . they cause what is known as oxidative stress in the cell by reacting with and damaging dna and lipid-based molecules like cell membranes . these ross are the reason we need antioxidants in our diet . mitochondria just like a factory can ’ t run without electricity , a cell can ’ t run without energy . atp ( adenosine triphosphate ) is the energy currency of the cell , and is produced in a process known as cellular respiration . though the process begins in the cytoplasm , the bulk of the energy produced comes from later steps that take place in the mitochondria . like we saw with the nuclear envelope , there are actually two lipid bilayers that separate the mitochondrial contents from the cytoplasm . we refer to them as the inner and outer mitochondrial membranes . if we cross both membranes we end up in the matrix , where pyruvate is sent after it is created from the breakdown of glucose ( this is step 1 of cellular respiration , known as glycolysis ) .the space between the two membranes is called the intermembrane space , and it has a low ph ( is acidic ) because the electron transport chain embedded in the inner membrane pumps protons ( h+ ) into it . energy to make atp comes from protons moving back into the matrix down their gradient from the intermembrane space . mitochondria are also somewhat unique in that they are self-replicating and have their own dna , almost as if they were a completely separate cell . the prevailing theory , known as the endosymbiotic theory , is that eukaryotes were first formed by large prokaryotic cells engulfing smaller cells that looked a lot like mitochondria ( and chloroplasts , more on them later ) . instead of being digested , the engulfed cells remained intact and the arrangement turned out to be advantageous to both cells , which created a symbiotic relationship . so far we ’ ve discussed organelles , the membrane-bound structures within a cell that have some sort of specialized function . now let ’ s take a moment to talk about the scaffolding that ’ s holding all of this in place - the walls and beams of our factory . cytoskeleton within the cytoplasm there is network of protein fibers known as the cytoskeleton . this structure is responsible for both cell movement and stability . the major components of the cytoskeleton are microtubules , intermediate filaments , and microfilaments . microtubules microtubules are small tubes made from the protein tubulin . these tubules are found in cilia and flagella , structures involved in cell movement . they also help provide pathways for secretory vesicles to move through the cell , and are even involved in cell division as they are a part of the mitotic spindle , which pulls homologous chromosomes apart . intermediate filaments smaller than the microtubules , but larger than the microfilaments , the intermediate filaments are made of a variety of proteins such as keratin and/or neurofilament . they are very stable , and help provide structure to the nuclear envelope and anchor organelles . microfilaments microfilaments are the thinnest part of the cytoskeleton , and are made of actin [ a highly-conserved protein that is actually the most abundant protein in most eukaryotic cells ] . actin is both flexible and strong , making it a useful protein in cell movement . in the heart , contraction is mediated through an actin-myosin system . plants and platelets so far we ’ ve covered basic organelles found in a eukaryotic cell . however , not every cell has each of these organelles , and some cells have organelles we haven ’ t discussed . for example , plant cells have chloroplasts , organelles that resemble mitochondria and are responsible for turning sunlight into useful energy for the cell ( this is like factories that are powered by energy they collect via solar panels ) . on the other hand , platelets , blood cells responsible for clotting , have no nucleus and are in fact just fragments of cytoplasm contained within a cell membrane . eukaryotes vs bacteria vs archaea it is also important to keep in mind that organelles are found only in eukaryotes , one of the three major cell divisions . the other two major divisions , bacteria and archaea are known as prokaryotes , and have no membrane bound organelles within . consider the following : some diseases can be traced back to organelle lack / malformation . for example , inclusion-cell ( i-cell ) disease occurs due to a defect in the golgi . in order to mark enzymes that should be sent to lysosomes to help degrade unwanted molecules , the golgi has to bind them with a mannose 6-phosphate tag , like a shipping label . however , in patients with i-cell disease , one of the proteins that make this tag is mutated , and can not do its job , like a broken label machine . this means that proteins can not be targeted to lysosomes . these untagged proteins are the enzymes that are responsible for chopping up other proteins . what happens is the inactivated enzymes end up being sent outside the cell , while lysosomes clog up with undigested material . this disease is congenital , and usually fatal before patients reach 7 years of age . an interesting idea is that mitochondria can be used to trace maternal ancestry . since mitochondria are self-replicating and have their own dna , they are not determined by the genes found in the nucleus . instead , your mitochondria have developed from the mitochondria present in the female ovum ( egg ) that you developed from . defects in mitochondrial dna cause hereditary diseases that pass only from mother to children .
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mitochondria are also somewhat unique in that they are self-replicating and have their own dna , almost as if they were a completely separate cell . the prevailing theory , known as the endosymbiotic theory , is that eukaryotes were first formed by large prokaryotic cells engulfing smaller cells that looked a lot like mitochondria ( and chloroplasts , more on them later ) . instead of being digested , the engulfed cells remained intact and the arrangement turned out to be advantageous to both cells , which created a symbiotic relationship .
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if the endosymbiotic theory is true then how did eukaryotes come to have nuclei ?
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not just any sky landscape paintings are often meant to document the look of a particular time in a particular place , to freeze a single moment and preserve it for eternity . el greco ’ s view of toledo does not do that . although the large church is placed in the correct place in the city , el greco changed the locations of several other buildings , proving that documentation was not the artist ’ s primary concern . rather than telling us what toledo looked like , here , el greco communicates what the city feels like . toledo becomes the means through which the artist expresses an interior psychological state , and perhaps , a view about the nature of man ’ s relationship with the divine . using typically dark , moody colors , el greco presented the spanish city of toledo at the top of a rolling hill . the city itself takes up only a little space in the center of the painting . the landscape and sky dominate . this is not just any sky . el greco ’ s clouds are about to crack open and unleash a storm on the city . the buildings themselves seem to crawl across the painting , and curving lines throughout the hill give the impression that the vista is moving , that it might actually be alive . in el greco ’ s toledo , something is about to happen , and it probably isn ’ t going to be good . something new : a cityscape to understand how radical this painting is , we have to weigh a few historical circumstances . first , el greco was painting in counter reformation spain , where religious dictates based on the council of trent ( which ended in 1563 ) , banned the landscape as a suitable subject for painting . although the church was his primary patron , the artist broke with that convention , and because of this , view of toledo has been called the first spanish landscape . more impressively , cityscapes never existed anywhere in the sixteenth century . el greco may literally have invented the genre . some art historians found this so unsettling that they had suggested that , because el greco often included views of toledo in the backgrounds of his religious paintings and portraits , view of toledo may have actually been cut from the background of a larger painting . however , we now know that this is not true . the city of toledo although el greco , “ the greek , ” is most usually known as a spanish painter , he was born domenikos theotokopouolos in crete in 1541 , and spent much of his life in italy . he was trained in the tradition of byzantine icon paintings in either crete or venice , where many cretans had settled , and by the 1560s was painting in titian ’ s workshop . in the 1570s he went to rome . although el greco was well reputed in italy , he failed to secure any commissions in the city , and was convinced by a spaniard to move to toledo , where he spent the next forty years of his life , and where he died in 1614 . why did the city of toledo inspired el greco to paint such a powerful picture of the city ? in spain , el greco failed to find favor with the king , and instead worked for the catholic church . if he was not raised in the faith , he almost certainly would have had to convert to catholicism . in the 1500s , spain ’ s catholic church had undergone huge transformations . the century started with the spanish inquisition , in which non-catholics were hunted out , tried , tortured , and often , killed . at the same time , people , like saint theresa of avilla and saint ignatius of loyola ( both spanish ) , were preaching that , through prayer , one could be directly inspired by god , and they claimed to have frequent visions in which god spoke to them . because of their beliefs , even these saints came under the scrutiny of the inquisition , although they were eventually acquitted . spain ’ s brand of catholicism , compared to italy ’ s , was mystical and based on personal experience . mysticism and inner conflict this mysticism is reflected in el greco ’ s view of toledo . almost entirely subsumed by the landscape , the city seems to be at the direct mercy of god . this is not a forgiving god , but rather a wrathful one , as in the old testament . toledo is undergoing a reckoning . at the same time , the landscape transcends this religious reading . it becomes reflective of the inner conflict of each human being , the feeling that making one ’ s way in the world is a harrowing endeavor . view of toledo was centuries ahead of its time , and probably can best be compared to vincent van gogh ’ s starry night , 1889 , which contains many of the same compositional elements ( a swirling sky , overwhelming nature , a skyline dominated by a church ) . but whereas van gogh evokes the calm of a little sleeping town , el greco ’ s painting captures the violence of the exterior world against an interior one . in this way , view of toledo has much in common with giorgione ’ s the tempest , in which a crack of lightning and oncoming storm threaten a woman and child seated in the landscape . el greco reminds us that there is an unfriendly world outside of us and that we are all subject to forces beyond our control . he leaves it up to us to decide whether we will succumb or prevail . text by christine zappella
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toledo is undergoing a reckoning . at the same time , the landscape transcends this religious reading . it becomes reflective of the inner conflict of each human being , the feeling that making one ’ s way in the world is a harrowing endeavor .
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was it just that they thought that only explicitly religious subjects should be painted ?
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not just any sky landscape paintings are often meant to document the look of a particular time in a particular place , to freeze a single moment and preserve it for eternity . el greco ’ s view of toledo does not do that . although the large church is placed in the correct place in the city , el greco changed the locations of several other buildings , proving that documentation was not the artist ’ s primary concern . rather than telling us what toledo looked like , here , el greco communicates what the city feels like . toledo becomes the means through which the artist expresses an interior psychological state , and perhaps , a view about the nature of man ’ s relationship with the divine . using typically dark , moody colors , el greco presented the spanish city of toledo at the top of a rolling hill . the city itself takes up only a little space in the center of the painting . the landscape and sky dominate . this is not just any sky . el greco ’ s clouds are about to crack open and unleash a storm on the city . the buildings themselves seem to crawl across the painting , and curving lines throughout the hill give the impression that the vista is moving , that it might actually be alive . in el greco ’ s toledo , something is about to happen , and it probably isn ’ t going to be good . something new : a cityscape to understand how radical this painting is , we have to weigh a few historical circumstances . first , el greco was painting in counter reformation spain , where religious dictates based on the council of trent ( which ended in 1563 ) , banned the landscape as a suitable subject for painting . although the church was his primary patron , the artist broke with that convention , and because of this , view of toledo has been called the first spanish landscape . more impressively , cityscapes never existed anywhere in the sixteenth century . el greco may literally have invented the genre . some art historians found this so unsettling that they had suggested that , because el greco often included views of toledo in the backgrounds of his religious paintings and portraits , view of toledo may have actually been cut from the background of a larger painting . however , we now know that this is not true . the city of toledo although el greco , “ the greek , ” is most usually known as a spanish painter , he was born domenikos theotokopouolos in crete in 1541 , and spent much of his life in italy . he was trained in the tradition of byzantine icon paintings in either crete or venice , where many cretans had settled , and by the 1560s was painting in titian ’ s workshop . in the 1570s he went to rome . although el greco was well reputed in italy , he failed to secure any commissions in the city , and was convinced by a spaniard to move to toledo , where he spent the next forty years of his life , and where he died in 1614 . why did the city of toledo inspired el greco to paint such a powerful picture of the city ? in spain , el greco failed to find favor with the king , and instead worked for the catholic church . if he was not raised in the faith , he almost certainly would have had to convert to catholicism . in the 1500s , spain ’ s catholic church had undergone huge transformations . the century started with the spanish inquisition , in which non-catholics were hunted out , tried , tortured , and often , killed . at the same time , people , like saint theresa of avilla and saint ignatius of loyola ( both spanish ) , were preaching that , through prayer , one could be directly inspired by god , and they claimed to have frequent visions in which god spoke to them . because of their beliefs , even these saints came under the scrutiny of the inquisition , although they were eventually acquitted . spain ’ s brand of catholicism , compared to italy ’ s , was mystical and based on personal experience . mysticism and inner conflict this mysticism is reflected in el greco ’ s view of toledo . almost entirely subsumed by the landscape , the city seems to be at the direct mercy of god . this is not a forgiving god , but rather a wrathful one , as in the old testament . toledo is undergoing a reckoning . at the same time , the landscape transcends this religious reading . it becomes reflective of the inner conflict of each human being , the feeling that making one ’ s way in the world is a harrowing endeavor . view of toledo was centuries ahead of its time , and probably can best be compared to vincent van gogh ’ s starry night , 1889 , which contains many of the same compositional elements ( a swirling sky , overwhelming nature , a skyline dominated by a church ) . but whereas van gogh evokes the calm of a little sleeping town , el greco ’ s painting captures the violence of the exterior world against an interior one . in this way , view of toledo has much in common with giorgione ’ s the tempest , in which a crack of lightning and oncoming storm threaten a woman and child seated in the landscape . el greco reminds us that there is an unfriendly world outside of us and that we are all subject to forces beyond our control . he leaves it up to us to decide whether we will succumb or prevail . text by christine zappella
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more impressively , cityscapes never existed anywhere in the sixteenth century . el greco may literally have invented the genre . some art historians found this so unsettling that they had suggested that , because el greco often included views of toledo in the backgrounds of his religious paintings and portraits , view of toledo may have actually been cut from the background of a larger painting .
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how are we so certain that `` el greco '' even produced these works in this time frame ?
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not just any sky landscape paintings are often meant to document the look of a particular time in a particular place , to freeze a single moment and preserve it for eternity . el greco ’ s view of toledo does not do that . although the large church is placed in the correct place in the city , el greco changed the locations of several other buildings , proving that documentation was not the artist ’ s primary concern . rather than telling us what toledo looked like , here , el greco communicates what the city feels like . toledo becomes the means through which the artist expresses an interior psychological state , and perhaps , a view about the nature of man ’ s relationship with the divine . using typically dark , moody colors , el greco presented the spanish city of toledo at the top of a rolling hill . the city itself takes up only a little space in the center of the painting . the landscape and sky dominate . this is not just any sky . el greco ’ s clouds are about to crack open and unleash a storm on the city . the buildings themselves seem to crawl across the painting , and curving lines throughout the hill give the impression that the vista is moving , that it might actually be alive . in el greco ’ s toledo , something is about to happen , and it probably isn ’ t going to be good . something new : a cityscape to understand how radical this painting is , we have to weigh a few historical circumstances . first , el greco was painting in counter reformation spain , where religious dictates based on the council of trent ( which ended in 1563 ) , banned the landscape as a suitable subject for painting . although the church was his primary patron , the artist broke with that convention , and because of this , view of toledo has been called the first spanish landscape . more impressively , cityscapes never existed anywhere in the sixteenth century . el greco may literally have invented the genre . some art historians found this so unsettling that they had suggested that , because el greco often included views of toledo in the backgrounds of his religious paintings and portraits , view of toledo may have actually been cut from the background of a larger painting . however , we now know that this is not true . the city of toledo although el greco , “ the greek , ” is most usually known as a spanish painter , he was born domenikos theotokopouolos in crete in 1541 , and spent much of his life in italy . he was trained in the tradition of byzantine icon paintings in either crete or venice , where many cretans had settled , and by the 1560s was painting in titian ’ s workshop . in the 1570s he went to rome . although el greco was well reputed in italy , he failed to secure any commissions in the city , and was convinced by a spaniard to move to toledo , where he spent the next forty years of his life , and where he died in 1614 . why did the city of toledo inspired el greco to paint such a powerful picture of the city ? in spain , el greco failed to find favor with the king , and instead worked for the catholic church . if he was not raised in the faith , he almost certainly would have had to convert to catholicism . in the 1500s , spain ’ s catholic church had undergone huge transformations . the century started with the spanish inquisition , in which non-catholics were hunted out , tried , tortured , and often , killed . at the same time , people , like saint theresa of avilla and saint ignatius of loyola ( both spanish ) , were preaching that , through prayer , one could be directly inspired by god , and they claimed to have frequent visions in which god spoke to them . because of their beliefs , even these saints came under the scrutiny of the inquisition , although they were eventually acquitted . spain ’ s brand of catholicism , compared to italy ’ s , was mystical and based on personal experience . mysticism and inner conflict this mysticism is reflected in el greco ’ s view of toledo . almost entirely subsumed by the landscape , the city seems to be at the direct mercy of god . this is not a forgiving god , but rather a wrathful one , as in the old testament . toledo is undergoing a reckoning . at the same time , the landscape transcends this religious reading . it becomes reflective of the inner conflict of each human being , the feeling that making one ’ s way in the world is a harrowing endeavor . view of toledo was centuries ahead of its time , and probably can best be compared to vincent van gogh ’ s starry night , 1889 , which contains many of the same compositional elements ( a swirling sky , overwhelming nature , a skyline dominated by a church ) . but whereas van gogh evokes the calm of a little sleeping town , el greco ’ s painting captures the violence of the exterior world against an interior one . in this way , view of toledo has much in common with giorgione ’ s the tempest , in which a crack of lightning and oncoming storm threaten a woman and child seated in the landscape . el greco reminds us that there is an unfriendly world outside of us and that we are all subject to forces beyond our control . he leaves it up to us to decide whether we will succumb or prevail . text by christine zappella
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this is not just any sky . el greco ’ s clouds are about to crack open and unleash a storm on the city . the buildings themselves seem to crawl across the painting , and curving lines throughout the hill give the impression that the vista is moving , that it might actually be alive .
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has n't recent information concerning the inquisition -- specifically , the vatican 's more open records , since 2000 -- caused a bit of a volte-face over the popular historical opinion dubbed the black legend ?
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not just any sky landscape paintings are often meant to document the look of a particular time in a particular place , to freeze a single moment and preserve it for eternity . el greco ’ s view of toledo does not do that . although the large church is placed in the correct place in the city , el greco changed the locations of several other buildings , proving that documentation was not the artist ’ s primary concern . rather than telling us what toledo looked like , here , el greco communicates what the city feels like . toledo becomes the means through which the artist expresses an interior psychological state , and perhaps , a view about the nature of man ’ s relationship with the divine . using typically dark , moody colors , el greco presented the spanish city of toledo at the top of a rolling hill . the city itself takes up only a little space in the center of the painting . the landscape and sky dominate . this is not just any sky . el greco ’ s clouds are about to crack open and unleash a storm on the city . the buildings themselves seem to crawl across the painting , and curving lines throughout the hill give the impression that the vista is moving , that it might actually be alive . in el greco ’ s toledo , something is about to happen , and it probably isn ’ t going to be good . something new : a cityscape to understand how radical this painting is , we have to weigh a few historical circumstances . first , el greco was painting in counter reformation spain , where religious dictates based on the council of trent ( which ended in 1563 ) , banned the landscape as a suitable subject for painting . although the church was his primary patron , the artist broke with that convention , and because of this , view of toledo has been called the first spanish landscape . more impressively , cityscapes never existed anywhere in the sixteenth century . el greco may literally have invented the genre . some art historians found this so unsettling that they had suggested that , because el greco often included views of toledo in the backgrounds of his religious paintings and portraits , view of toledo may have actually been cut from the background of a larger painting . however , we now know that this is not true . the city of toledo although el greco , “ the greek , ” is most usually known as a spanish painter , he was born domenikos theotokopouolos in crete in 1541 , and spent much of his life in italy . he was trained in the tradition of byzantine icon paintings in either crete or venice , where many cretans had settled , and by the 1560s was painting in titian ’ s workshop . in the 1570s he went to rome . although el greco was well reputed in italy , he failed to secure any commissions in the city , and was convinced by a spaniard to move to toledo , where he spent the next forty years of his life , and where he died in 1614 . why did the city of toledo inspired el greco to paint such a powerful picture of the city ? in spain , el greco failed to find favor with the king , and instead worked for the catholic church . if he was not raised in the faith , he almost certainly would have had to convert to catholicism . in the 1500s , spain ’ s catholic church had undergone huge transformations . the century started with the spanish inquisition , in which non-catholics were hunted out , tried , tortured , and often , killed . at the same time , people , like saint theresa of avilla and saint ignatius of loyola ( both spanish ) , were preaching that , through prayer , one could be directly inspired by god , and they claimed to have frequent visions in which god spoke to them . because of their beliefs , even these saints came under the scrutiny of the inquisition , although they were eventually acquitted . spain ’ s brand of catholicism , compared to italy ’ s , was mystical and based on personal experience . mysticism and inner conflict this mysticism is reflected in el greco ’ s view of toledo . almost entirely subsumed by the landscape , the city seems to be at the direct mercy of god . this is not a forgiving god , but rather a wrathful one , as in the old testament . toledo is undergoing a reckoning . at the same time , the landscape transcends this religious reading . it becomes reflective of the inner conflict of each human being , the feeling that making one ’ s way in the world is a harrowing endeavor . view of toledo was centuries ahead of its time , and probably can best be compared to vincent van gogh ’ s starry night , 1889 , which contains many of the same compositional elements ( a swirling sky , overwhelming nature , a skyline dominated by a church ) . but whereas van gogh evokes the calm of a little sleeping town , el greco ’ s painting captures the violence of the exterior world against an interior one . in this way , view of toledo has much in common with giorgione ’ s the tempest , in which a crack of lightning and oncoming storm threaten a woman and child seated in the landscape . el greco reminds us that there is an unfriendly world outside of us and that we are all subject to forces beyond our control . he leaves it up to us to decide whether we will succumb or prevail . text by christine zappella
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el greco may literally have invented the genre . some art historians found this so unsettling that they had suggested that , because el greco often included views of toledo in the backgrounds of his religious paintings and portraits , view of toledo may have actually been cut from the background of a larger painting . however , we now know that this is not true .
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i 'm not certain how widely historians , and art historians , regard this to be true , but how might it change your interpretation of this painting ?
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not just any sky landscape paintings are often meant to document the look of a particular time in a particular place , to freeze a single moment and preserve it for eternity . el greco ’ s view of toledo does not do that . although the large church is placed in the correct place in the city , el greco changed the locations of several other buildings , proving that documentation was not the artist ’ s primary concern . rather than telling us what toledo looked like , here , el greco communicates what the city feels like . toledo becomes the means through which the artist expresses an interior psychological state , and perhaps , a view about the nature of man ’ s relationship with the divine . using typically dark , moody colors , el greco presented the spanish city of toledo at the top of a rolling hill . the city itself takes up only a little space in the center of the painting . the landscape and sky dominate . this is not just any sky . el greco ’ s clouds are about to crack open and unleash a storm on the city . the buildings themselves seem to crawl across the painting , and curving lines throughout the hill give the impression that the vista is moving , that it might actually be alive . in el greco ’ s toledo , something is about to happen , and it probably isn ’ t going to be good . something new : a cityscape to understand how radical this painting is , we have to weigh a few historical circumstances . first , el greco was painting in counter reformation spain , where religious dictates based on the council of trent ( which ended in 1563 ) , banned the landscape as a suitable subject for painting . although the church was his primary patron , the artist broke with that convention , and because of this , view of toledo has been called the first spanish landscape . more impressively , cityscapes never existed anywhere in the sixteenth century . el greco may literally have invented the genre . some art historians found this so unsettling that they had suggested that , because el greco often included views of toledo in the backgrounds of his religious paintings and portraits , view of toledo may have actually been cut from the background of a larger painting . however , we now know that this is not true . the city of toledo although el greco , “ the greek , ” is most usually known as a spanish painter , he was born domenikos theotokopouolos in crete in 1541 , and spent much of his life in italy . he was trained in the tradition of byzantine icon paintings in either crete or venice , where many cretans had settled , and by the 1560s was painting in titian ’ s workshop . in the 1570s he went to rome . although el greco was well reputed in italy , he failed to secure any commissions in the city , and was convinced by a spaniard to move to toledo , where he spent the next forty years of his life , and where he died in 1614 . why did the city of toledo inspired el greco to paint such a powerful picture of the city ? in spain , el greco failed to find favor with the king , and instead worked for the catholic church . if he was not raised in the faith , he almost certainly would have had to convert to catholicism . in the 1500s , spain ’ s catholic church had undergone huge transformations . the century started with the spanish inquisition , in which non-catholics were hunted out , tried , tortured , and often , killed . at the same time , people , like saint theresa of avilla and saint ignatius of loyola ( both spanish ) , were preaching that , through prayer , one could be directly inspired by god , and they claimed to have frequent visions in which god spoke to them . because of their beliefs , even these saints came under the scrutiny of the inquisition , although they were eventually acquitted . spain ’ s brand of catholicism , compared to italy ’ s , was mystical and based on personal experience . mysticism and inner conflict this mysticism is reflected in el greco ’ s view of toledo . almost entirely subsumed by the landscape , the city seems to be at the direct mercy of god . this is not a forgiving god , but rather a wrathful one , as in the old testament . toledo is undergoing a reckoning . at the same time , the landscape transcends this religious reading . it becomes reflective of the inner conflict of each human being , the feeling that making one ’ s way in the world is a harrowing endeavor . view of toledo was centuries ahead of its time , and probably can best be compared to vincent van gogh ’ s starry night , 1889 , which contains many of the same compositional elements ( a swirling sky , overwhelming nature , a skyline dominated by a church ) . but whereas van gogh evokes the calm of a little sleeping town , el greco ’ s painting captures the violence of the exterior world against an interior one . in this way , view of toledo has much in common with giorgione ’ s the tempest , in which a crack of lightning and oncoming storm threaten a woman and child seated in the landscape . el greco reminds us that there is an unfriendly world outside of us and that we are all subject to forces beyond our control . he leaves it up to us to decide whether we will succumb or prevail . text by christine zappella
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this is not just any sky . el greco ’ s clouds are about to crack open and unleash a storm on the city . the buildings themselves seem to crawl across the painting , and curving lines throughout the hill give the impression that the vista is moving , that it might actually be alive .
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it is n't an allusion that she is depicted in the clouds is it ?
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the roman state religion in microcosm the festivities of the roman state religion were steeped in tradition and ritual symbolism . sacred offerings to the gods , consultations with priests and diviners , ritual formulae , communal feasting—were all practices aimed at fostering and maintaining social cohesion and communicating authority . it could perhaps be argued that the ara pacis augustae—the altar of augustan peace—represents in luxurious , stately microcosm the practices of the roman state religion in a way that is simultaneously elegant and pragmatic . vowed on july 4 , 13 b.c.e. , and dedicated on january 30 , 9 b.c.e. , the monument stood proudly in the campus martius in rome ( a level area between several of rome 's hills and the tiber river ) . it was adjacent to architectural complexes that cultivated and proudly displayed messages about the power , legitimacy , and suitability of their patron—the emperor augustus . now excavated , restored , and reassembled in a sleek modern pavilion designed by architect richard meier ( 2006 ) , the ara pacis continues to inspire and challenge us as we think about ancient rome . augustus himself discusses the ara pacis in his epigraphical memoir , res gestae divi augusti ( “ deeds of the divine augustus ” ) that was promulgated upon his death in 14 c.e . augustus states “ when i returned to rome from spain and gaul , having successfully accomplished deeds in those provinces … the senate voted to consecrate the altar of august peace in the campus martius … on which it ordered the magistrates and priests and vestal virgins to offer annual sacrifices ” ( aug. rg 12 ) . an open-air altar for sacrifice the ara pacis is , at its simplest , an open-air altar for blood sacrifice associated with the roman state religion . the ritual slaughtering and offering of animals in roman religion was routine , and such rites usually took place outdoors . the placement of the ara pacis in the campus martius ( field of mars ) along the via lata ( now the via del corso ) situated it close to other key augustan monuments , notably the horologium augusti ( a giant sundial ) and the mausoleum of augustus . the significance of the topographical placement would have been quite evident to ancient romans . this complex of augustan monuments made a clear statement about augustus ’ physical transformation of rome ’ s urban landscape . the dedication to a rather abstract notion of peace ( pax ) is significant in that augustus advertises the fact that he has restored peace to the roman state after a long period of internal and external turmoil . the altar ( ara ) itself sits within a monumental stone screen that has been elaborated with bas relief ( low relief ) sculpture , with the panels combining to form a programmatic mytho-historical narrative about augustus and his administration , as well as about rome ’ s deep roots . the altar enclosure is roughly square while the altar itself sits atop a raised podium that is accessible via a narrow stairway . the outer screen—processional scenes processional scenes occupy the north and south flanks of the altar screen . the solemn figures , all properly clad for a rite of the state religion , proceed in the direction of the altar itself , ready to participate in the ritual . the figures all advance toward the west . the occasion depicted would seem to be a celebration of the peace ( pax ) that augustus had restored to the roman empire . in addition four main groups of people are evident in the processions : ( 1 ) the lictors ( the official bodyguards of magistrates ) , ( 2 ) priests from the major collegia of rome , ( 3 ) members of the imperial household , including women and children , and ( 4 ) attendants . there has been a good deal of scholarly discussion focused on two of three non-roman children who are depicted . the north processional frieze , made up of priests and members of the imperial household , is comprised of 46 figures . the priestly colleges ( religious associations ) represented include the septemviri epulones ( `` seven men for sacrificial banquets '' —they arranged public feasts connected to sacred holidays ) , whose members here carry an incense box ( image above ) , and the quindecimviri sacris faciundis ( `` fifteen men to perform sacred actions '' — their main duty was to guard and consult the sibylline books ( oracular texts ) at the request of the senate ) . members of the imperial family , including octavia minor , follow behind . a good deal of modern restoration has been undertaken on the north wall , with many heads heavily restored or replaced . the south wall of the exterior screen depicts augustus and his immediate family . the identification of the individual figures has been the source of a great deal of scholarly debate . depicted here are augustus ( damaged , he appears at the far left in the image above ) and marcus agrippa ( friend , son-in-law , and lieutenant to augustus , he appears , hooded , image below ) , along with other members of the imperial house . all of those present are dressed in ceremonial garb appropriate for the state sacrifice . the presence of state priests known as flamens ( flamines ) further indicate the solemnity of the occasion . a running , vegetal frieze runs parallel to the processional friezes on the lower register . this vegetal frieze emphasizes the fertility and abundance of the lands , a clear benefit of living in a time of peace . mythological panels accompanying the processional friezes are four mythological panels that adorn the altar screen on its shorter sides . each of these panels depicts a distinct scene : a scene of a bearded male making sacrifice ( below ) a scene of seated female goddess amid the fertility of italy ( also below ) a fragmentary scene with romulus and remus in the lupercal grotto ( where these two mythic founders of rome were suckled by a she-wolf ) and a fragmentary panel showing roma ( the personification of rome ) as a seated goddess . since the early twentieth century , the mainstream interpretation of the sacrifice panel ( above ) has been that the scene depicts the trojan hero aeneas arriving in italy and making a sacrifice to juno . a recent re-interpretation offered by paul rehak argues instead that the bearded man is not aeneas , but numa pompilius , rome ’ s second king . in rehak ’ s theory , numa , renowned as a peaceful ruler and the founder of roman religion , provides a counterbalance to the warlike romulus on the opposite panel . the better preserved panel of the east wall depicts a seated female figure ( above ) who has been variously interpreted as tellus ( the earth ) , italia ( italy ) , pax ( peace ) , as well as venus . the panel depicts a scene of human fertility and natural abundance . two babies sit on the lap of the seated female , tugging at her drapery . surrounding the central female is the natural abundance of the lands and flanking her are the personifications of the land and sea breezes . in all , whether the goddess is taken as tellus or pax , the theme stressed is the harmony and abundance of italy , a theme central to augustus ’ message of a restored peaceful state for the roman people—the pax romana . the altar the altar itself ( below ) sits within the sculpted precinct wall . it is framed by sculpted architectural mouldings with crouching gryphons surmounted by volutes flanking the altar . the altar was the functional portion of the monument , the place where blood sacrifice and/or burnt offerings would be presented to the gods . implications and interpretation the implications of the ara pacis are far reaching . originally located along the via lata ( now rome ’ s via del corso ) , the altar is part of a monumental architectural makeover of rome ’ s campus martius carried out by augustus and his family . initially the makeover had a dynastic tone , with the mausoleum of augustus near the river . the dedication of the horologium ( sundial ) of augustus and the ara pacis , the augustan makeover served as a potent , visual reminder of augustus ’ success to the people of rome . the choice to celebrate peace and the attendant prosperity in some ways breaks with the tradition of explicitly triumphal monuments that advertise success in war and victories won on the battlefield . by championing peace—at least in the guise of public monuments—augustus promoted a powerful and effective campaign of political message making . rediscovery the first fragments of the ara pacis emerged in 1568 beneath rome ’ s palazzo chigi near the basilica of san lorenzo in lucina . these initial fragments came to be dispersed among various museums , including the villa medici , the vatican museums , the louvre , and the uffizi . it was not until 1859 that further fragments of the ara pacis emerged . the german art historian friedrich von duhn of the university of heidelberg is credited with the discovery that the fragments corresponded to the altar mentioned in augustus ’ res gestae . although von duhn reached this conclusion by 1881 , excavations were not resumed until 1903 , at which time the total number of recovered fragments reached 53 , after which the excavation was again halted due to difficult conditions . work at the site began again in february 1937 when advanced technology was used to freeze approximately 70 cubic meters of soil to allow for the extraction of the remaining fragments . this excavation was mandated by the order of the italian government of benito mussolini and his planned jubilee in 1938 that was designed to commemorate the 2,000th anniversary of augustus ’ birth . mussolini and augustus the revival of the glory of ancient rome was central to the propaganda of the fascist regime in italy during the 1930s . benito mussolini himself cultivated a connection with the personage of augustus and claimed his actions were aimed at furthering the continuity of the roman empire . art , architecture , and iconography played a key role in this propagandistic “ revival ” . following the 1937 retrieval of additional fragments of the altar , mussolini directed architect vittorio ballio morpurgo to construct an enclosure for the restored altar adjacent to the ruins of the mausoleum of augustus near the tiber river , creating a key complex for fascist propaganda . newly built fascist palaces , bearing fascist propaganda , flank the space dubbed “ piazza augusto imperatore ” ( “ plaza of the emperor augustus ” ) . the famous res gestae divi augusti ( “ deeds of the divine augustus ” ) was re-created on the wall of the altar ’ s pavilion . the concomitant effect was meant to lead the viewer to associate mussolini ’ s accomplishments with those of augustus himself . the ara pacis and richard meier the firm of architect richard meier was engaged to design and execute a new and improved pavilion to house the ara pacis and to integrate the altar with a planned pedestrian area surrounding the adjacent mausoleum of augustus . between 1995 and the dedication of the new pavilion in 2006 meier crafted the modernist pavilion that capitalizes on glass curtain walls granting visitors views of the tiber river and the mausoleum while they perambulate in the museum space focused on the altar itself . the meier pavilion has not been well-received , with some critics immediately panning it and some italian politicians declaring that it should be dismantled . the museum has also been the victim of targeted vandalism . enduring monumentality the ara pacis augustae continues to engage us and to incite controversy . as a monument that is the product of a carefully constructed ideological program , it is highly charged with socio-cultural energy that speaks to us about the ordering of the roman world and its society—the very roman universe . augustus had a strong interest in reshaping the roman world ( with him as the sole leader ) , but had to be cautious about how radical those changes seemed to the roman populace . while he defeated enemies , both foreign and domestic , he was concerned about being perceived as too authoritarian -- he did not wish to labeled as a king ( rex ) for fear that this would be too much for the roman people to accept . so the augustan scheme involved a declaration that rome 's republican government had been `` restored '' by augustus and he styled himself as the leading citizen of the republic ( princeps ) . these political and ideological motives then influence and guide the creation of his program of monumental art and architecture . these monumental forms , of which the ara pacis is a prime example , served to both create and reinforce these augustan messages . the story of the ara pacis become even more complicated since it is an artifact that then was placed in the service of ideas in the modern age . this results in its identity being impossibly , a mixture of classicism and fascism and modernism—all difficult to interpret in a postmodern reality . it is important to remember that the sculptural reliefs were created in the first place to be easily readable , so that the viewer could understand the messages of augustus and his circle without the need to read elaborate texts . augustus pioneered the use of such ideological messages that relied on clear iconography to get their message across . a great deal was at stake for augustus and it seems , by virtue of history , that the political choices he made proved prudent . the messages of the pax romana , of a restored state , and of augustus as a leading republican citizen , are all part of an effective and carefully constructed veneer . essay by dr. jeffrey a. becker additional resources : ara pacis museum ara pacis augustae ( reed college ) roma sparita photo archive ( italian ) ara pacis museum / richard meier & amp ; partners architects david castriota , the ara pacis augustae and the imagery of abundance in later greek and early roman imperial art ( princeton : princeton university press , 1995 ) . diane a. conlin , the artists of the ara pacis : the process of hellenization in roman relief sculpture ( chapel hill : university of north carolina press , 1997 ) . nancy de grummond , “ pax augusta and the horae on the ara pacis augustae , ” american journal of archaeology 94.4 ( 1990 ) pp . 663–677 . karl galinksy , augustan culture : an interpretive introduction ( princeton : princeton university press , 1996 ) . karl galinksy ed. , the cambridge companion to the age of augustus ( cambridge : cambridge university press , 2005 ) . peter heslin , `` augustus , domitian and the so-called horologium augusti , '' journal of roman studies , 97 ( 2007 ) , pp . 1-20 . p. j. holliday , “ time , history , and ritual on the ara pacis augustae , '' the art bulletin 72.4 ( december 1990 ) , pp . 542–557 . paul jacobs and diane conlin , campus martius : the field of mars in the life of ancient rome ( cambridge : cambridge university press , 2015 ) . diana e. e. kleiner , roman sculpture ( new haven : yale university press , 1994 ) . gerhard m. koeppel , `` the grand pictorial tradition of roman historical representation during the early empire , '' aufstieg und niedergang der römischen welt ii.12.1 ( 1982 ) , pp . 507-535 . gerhard m. koeppel , `` the role of pictorial models in the creation of the historical relief during the age of augustus , '' in the age of augustus , edited by r. winkes ( providence , r.i. : center for old world archaeology and art , brown university ; louvain-la-neuve , belgium : institut supérieur d'archéologie de d'histoire de l'art , collège érasme , 1985 ) , pp . 89-106 . paul rehak , “ aeneas or numa ? rethinking the meaning of the ara pacis augustae , ” the art bulletin 83.2 ( jun. , 2001 ) , pp . 190-208 . paul rehak , imperium and cosmos . augustus and the northern campus martius , edited by john g. younger . ( madison , wi : the university of wisconsin press , 2006 ) . alan riding , “ richard meier 's new home for the ara pacis , a roman treasure , opens , ” the new york times april 24 , 2006 . john seabrook , “ roman renovation , ” the new yorker , may 2 , 2005 pp . 56-65 . j. sieveking , “ zur ara pacis , ” jahresheft des österreichischen archeologischen institut 10 ( 1907 ) . catherine slessor , “ roman remains , ” architectural review , 219.1307 ( 2006 ) , pp . 18-19 . m. j. strazzulla , “ war and peace : housing the ara pacis in the eternal city , ” american journal of archaeology 113.2 ( 2009 ) pp . 1-10 . stefan weinstock , “ pax and the 'ara pacis ' , ” the journal of roman studies 50.1-2 ( 1960 ) pp . 44–58 . rolf winkes ed. , the age of augustus : interdisciplinary conference held at brown university , april 30-may 2 , 1982 ( providence , r.i. : center for old world archaeology and art , brown university ; louvain-la-neuve , belgium : institut supérieur d'archéologie de d'histoire de l'art , collège érasme , 1985 ) . paul zanker , the power of images in the age of augustus , trans . d. schneider ( ann arbor : university of michigan press , 1987 ) .
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the concomitant effect was meant to lead the viewer to associate mussolini ’ s accomplishments with those of augustus himself . the ara pacis and richard meier the firm of architect richard meier was engaged to design and execute a new and improved pavilion to house the ara pacis and to integrate the altar with a planned pedestrian area surrounding the adjacent mausoleum of augustus . between 1995 and the dedication of the new pavilion in 2006 meier crafted the modernist pavilion that capitalizes on glass curtain walls granting visitors views of the tiber river and the mausoleum while they perambulate in the museum space focused on the altar itself .
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is the `` architect richard meier '' the same `` meier '' of getty center los angeles fame ?
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the roman state religion in microcosm the festivities of the roman state religion were steeped in tradition and ritual symbolism . sacred offerings to the gods , consultations with priests and diviners , ritual formulae , communal feasting—were all practices aimed at fostering and maintaining social cohesion and communicating authority . it could perhaps be argued that the ara pacis augustae—the altar of augustan peace—represents in luxurious , stately microcosm the practices of the roman state religion in a way that is simultaneously elegant and pragmatic . vowed on july 4 , 13 b.c.e. , and dedicated on january 30 , 9 b.c.e. , the monument stood proudly in the campus martius in rome ( a level area between several of rome 's hills and the tiber river ) . it was adjacent to architectural complexes that cultivated and proudly displayed messages about the power , legitimacy , and suitability of their patron—the emperor augustus . now excavated , restored , and reassembled in a sleek modern pavilion designed by architect richard meier ( 2006 ) , the ara pacis continues to inspire and challenge us as we think about ancient rome . augustus himself discusses the ara pacis in his epigraphical memoir , res gestae divi augusti ( “ deeds of the divine augustus ” ) that was promulgated upon his death in 14 c.e . augustus states “ when i returned to rome from spain and gaul , having successfully accomplished deeds in those provinces … the senate voted to consecrate the altar of august peace in the campus martius … on which it ordered the magistrates and priests and vestal virgins to offer annual sacrifices ” ( aug. rg 12 ) . an open-air altar for sacrifice the ara pacis is , at its simplest , an open-air altar for blood sacrifice associated with the roman state religion . the ritual slaughtering and offering of animals in roman religion was routine , and such rites usually took place outdoors . the placement of the ara pacis in the campus martius ( field of mars ) along the via lata ( now the via del corso ) situated it close to other key augustan monuments , notably the horologium augusti ( a giant sundial ) and the mausoleum of augustus . the significance of the topographical placement would have been quite evident to ancient romans . this complex of augustan monuments made a clear statement about augustus ’ physical transformation of rome ’ s urban landscape . the dedication to a rather abstract notion of peace ( pax ) is significant in that augustus advertises the fact that he has restored peace to the roman state after a long period of internal and external turmoil . the altar ( ara ) itself sits within a monumental stone screen that has been elaborated with bas relief ( low relief ) sculpture , with the panels combining to form a programmatic mytho-historical narrative about augustus and his administration , as well as about rome ’ s deep roots . the altar enclosure is roughly square while the altar itself sits atop a raised podium that is accessible via a narrow stairway . the outer screen—processional scenes processional scenes occupy the north and south flanks of the altar screen . the solemn figures , all properly clad for a rite of the state religion , proceed in the direction of the altar itself , ready to participate in the ritual . the figures all advance toward the west . the occasion depicted would seem to be a celebration of the peace ( pax ) that augustus had restored to the roman empire . in addition four main groups of people are evident in the processions : ( 1 ) the lictors ( the official bodyguards of magistrates ) , ( 2 ) priests from the major collegia of rome , ( 3 ) members of the imperial household , including women and children , and ( 4 ) attendants . there has been a good deal of scholarly discussion focused on two of three non-roman children who are depicted . the north processional frieze , made up of priests and members of the imperial household , is comprised of 46 figures . the priestly colleges ( religious associations ) represented include the septemviri epulones ( `` seven men for sacrificial banquets '' —they arranged public feasts connected to sacred holidays ) , whose members here carry an incense box ( image above ) , and the quindecimviri sacris faciundis ( `` fifteen men to perform sacred actions '' — their main duty was to guard and consult the sibylline books ( oracular texts ) at the request of the senate ) . members of the imperial family , including octavia minor , follow behind . a good deal of modern restoration has been undertaken on the north wall , with many heads heavily restored or replaced . the south wall of the exterior screen depicts augustus and his immediate family . the identification of the individual figures has been the source of a great deal of scholarly debate . depicted here are augustus ( damaged , he appears at the far left in the image above ) and marcus agrippa ( friend , son-in-law , and lieutenant to augustus , he appears , hooded , image below ) , along with other members of the imperial house . all of those present are dressed in ceremonial garb appropriate for the state sacrifice . the presence of state priests known as flamens ( flamines ) further indicate the solemnity of the occasion . a running , vegetal frieze runs parallel to the processional friezes on the lower register . this vegetal frieze emphasizes the fertility and abundance of the lands , a clear benefit of living in a time of peace . mythological panels accompanying the processional friezes are four mythological panels that adorn the altar screen on its shorter sides . each of these panels depicts a distinct scene : a scene of a bearded male making sacrifice ( below ) a scene of seated female goddess amid the fertility of italy ( also below ) a fragmentary scene with romulus and remus in the lupercal grotto ( where these two mythic founders of rome were suckled by a she-wolf ) and a fragmentary panel showing roma ( the personification of rome ) as a seated goddess . since the early twentieth century , the mainstream interpretation of the sacrifice panel ( above ) has been that the scene depicts the trojan hero aeneas arriving in italy and making a sacrifice to juno . a recent re-interpretation offered by paul rehak argues instead that the bearded man is not aeneas , but numa pompilius , rome ’ s second king . in rehak ’ s theory , numa , renowned as a peaceful ruler and the founder of roman religion , provides a counterbalance to the warlike romulus on the opposite panel . the better preserved panel of the east wall depicts a seated female figure ( above ) who has been variously interpreted as tellus ( the earth ) , italia ( italy ) , pax ( peace ) , as well as venus . the panel depicts a scene of human fertility and natural abundance . two babies sit on the lap of the seated female , tugging at her drapery . surrounding the central female is the natural abundance of the lands and flanking her are the personifications of the land and sea breezes . in all , whether the goddess is taken as tellus or pax , the theme stressed is the harmony and abundance of italy , a theme central to augustus ’ message of a restored peaceful state for the roman people—the pax romana . the altar the altar itself ( below ) sits within the sculpted precinct wall . it is framed by sculpted architectural mouldings with crouching gryphons surmounted by volutes flanking the altar . the altar was the functional portion of the monument , the place where blood sacrifice and/or burnt offerings would be presented to the gods . implications and interpretation the implications of the ara pacis are far reaching . originally located along the via lata ( now rome ’ s via del corso ) , the altar is part of a monumental architectural makeover of rome ’ s campus martius carried out by augustus and his family . initially the makeover had a dynastic tone , with the mausoleum of augustus near the river . the dedication of the horologium ( sundial ) of augustus and the ara pacis , the augustan makeover served as a potent , visual reminder of augustus ’ success to the people of rome . the choice to celebrate peace and the attendant prosperity in some ways breaks with the tradition of explicitly triumphal monuments that advertise success in war and victories won on the battlefield . by championing peace—at least in the guise of public monuments—augustus promoted a powerful and effective campaign of political message making . rediscovery the first fragments of the ara pacis emerged in 1568 beneath rome ’ s palazzo chigi near the basilica of san lorenzo in lucina . these initial fragments came to be dispersed among various museums , including the villa medici , the vatican museums , the louvre , and the uffizi . it was not until 1859 that further fragments of the ara pacis emerged . the german art historian friedrich von duhn of the university of heidelberg is credited with the discovery that the fragments corresponded to the altar mentioned in augustus ’ res gestae . although von duhn reached this conclusion by 1881 , excavations were not resumed until 1903 , at which time the total number of recovered fragments reached 53 , after which the excavation was again halted due to difficult conditions . work at the site began again in february 1937 when advanced technology was used to freeze approximately 70 cubic meters of soil to allow for the extraction of the remaining fragments . this excavation was mandated by the order of the italian government of benito mussolini and his planned jubilee in 1938 that was designed to commemorate the 2,000th anniversary of augustus ’ birth . mussolini and augustus the revival of the glory of ancient rome was central to the propaganda of the fascist regime in italy during the 1930s . benito mussolini himself cultivated a connection with the personage of augustus and claimed his actions were aimed at furthering the continuity of the roman empire . art , architecture , and iconography played a key role in this propagandistic “ revival ” . following the 1937 retrieval of additional fragments of the altar , mussolini directed architect vittorio ballio morpurgo to construct an enclosure for the restored altar adjacent to the ruins of the mausoleum of augustus near the tiber river , creating a key complex for fascist propaganda . newly built fascist palaces , bearing fascist propaganda , flank the space dubbed “ piazza augusto imperatore ” ( “ plaza of the emperor augustus ” ) . the famous res gestae divi augusti ( “ deeds of the divine augustus ” ) was re-created on the wall of the altar ’ s pavilion . the concomitant effect was meant to lead the viewer to associate mussolini ’ s accomplishments with those of augustus himself . the ara pacis and richard meier the firm of architect richard meier was engaged to design and execute a new and improved pavilion to house the ara pacis and to integrate the altar with a planned pedestrian area surrounding the adjacent mausoleum of augustus . between 1995 and the dedication of the new pavilion in 2006 meier crafted the modernist pavilion that capitalizes on glass curtain walls granting visitors views of the tiber river and the mausoleum while they perambulate in the museum space focused on the altar itself . the meier pavilion has not been well-received , with some critics immediately panning it and some italian politicians declaring that it should be dismantled . the museum has also been the victim of targeted vandalism . enduring monumentality the ara pacis augustae continues to engage us and to incite controversy . as a monument that is the product of a carefully constructed ideological program , it is highly charged with socio-cultural energy that speaks to us about the ordering of the roman world and its society—the very roman universe . augustus had a strong interest in reshaping the roman world ( with him as the sole leader ) , but had to be cautious about how radical those changes seemed to the roman populace . while he defeated enemies , both foreign and domestic , he was concerned about being perceived as too authoritarian -- he did not wish to labeled as a king ( rex ) for fear that this would be too much for the roman people to accept . so the augustan scheme involved a declaration that rome 's republican government had been `` restored '' by augustus and he styled himself as the leading citizen of the republic ( princeps ) . these political and ideological motives then influence and guide the creation of his program of monumental art and architecture . these monumental forms , of which the ara pacis is a prime example , served to both create and reinforce these augustan messages . the story of the ara pacis become even more complicated since it is an artifact that then was placed in the service of ideas in the modern age . this results in its identity being impossibly , a mixture of classicism and fascism and modernism—all difficult to interpret in a postmodern reality . it is important to remember that the sculptural reliefs were created in the first place to be easily readable , so that the viewer could understand the messages of augustus and his circle without the need to read elaborate texts . augustus pioneered the use of such ideological messages that relied on clear iconography to get their message across . a great deal was at stake for augustus and it seems , by virtue of history , that the political choices he made proved prudent . the messages of the pax romana , of a restored state , and of augustus as a leading republican citizen , are all part of an effective and carefully constructed veneer . essay by dr. jeffrey a. becker additional resources : ara pacis museum ara pacis augustae ( reed college ) roma sparita photo archive ( italian ) ara pacis museum / richard meier & amp ; partners architects david castriota , the ara pacis augustae and the imagery of abundance in later greek and early roman imperial art ( princeton : princeton university press , 1995 ) . diane a. conlin , the artists of the ara pacis : the process of hellenization in roman relief sculpture ( chapel hill : university of north carolina press , 1997 ) . nancy de grummond , “ pax augusta and the horae on the ara pacis augustae , ” american journal of archaeology 94.4 ( 1990 ) pp . 663–677 . karl galinksy , augustan culture : an interpretive introduction ( princeton : princeton university press , 1996 ) . karl galinksy ed. , the cambridge companion to the age of augustus ( cambridge : cambridge university press , 2005 ) . peter heslin , `` augustus , domitian and the so-called horologium augusti , '' journal of roman studies , 97 ( 2007 ) , pp . 1-20 . p. j. holliday , “ time , history , and ritual on the ara pacis augustae , '' the art bulletin 72.4 ( december 1990 ) , pp . 542–557 . paul jacobs and diane conlin , campus martius : the field of mars in the life of ancient rome ( cambridge : cambridge university press , 2015 ) . diana e. e. kleiner , roman sculpture ( new haven : yale university press , 1994 ) . gerhard m. koeppel , `` the grand pictorial tradition of roman historical representation during the early empire , '' aufstieg und niedergang der römischen welt ii.12.1 ( 1982 ) , pp . 507-535 . gerhard m. koeppel , `` the role of pictorial models in the creation of the historical relief during the age of augustus , '' in the age of augustus , edited by r. winkes ( providence , r.i. : center for old world archaeology and art , brown university ; louvain-la-neuve , belgium : institut supérieur d'archéologie de d'histoire de l'art , collège érasme , 1985 ) , pp . 89-106 . paul rehak , “ aeneas or numa ? rethinking the meaning of the ara pacis augustae , ” the art bulletin 83.2 ( jun. , 2001 ) , pp . 190-208 . paul rehak , imperium and cosmos . augustus and the northern campus martius , edited by john g. younger . ( madison , wi : the university of wisconsin press , 2006 ) . alan riding , “ richard meier 's new home for the ara pacis , a roman treasure , opens , ” the new york times april 24 , 2006 . john seabrook , “ roman renovation , ” the new yorker , may 2 , 2005 pp . 56-65 . j. sieveking , “ zur ara pacis , ” jahresheft des österreichischen archeologischen institut 10 ( 1907 ) . catherine slessor , “ roman remains , ” architectural review , 219.1307 ( 2006 ) , pp . 18-19 . m. j. strazzulla , “ war and peace : housing the ara pacis in the eternal city , ” american journal of archaeology 113.2 ( 2009 ) pp . 1-10 . stefan weinstock , “ pax and the 'ara pacis ' , ” the journal of roman studies 50.1-2 ( 1960 ) pp . 44–58 . rolf winkes ed. , the age of augustus : interdisciplinary conference held at brown university , april 30-may 2 , 1982 ( providence , r.i. : center for old world archaeology and art , brown university ; louvain-la-neuve , belgium : institut supérieur d'archéologie de d'histoire de l'art , collège érasme , 1985 ) . paul zanker , the power of images in the age of augustus , trans . d. schneider ( ann arbor : university of michigan press , 1987 ) .
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the messages of the pax romana , of a restored state , and of augustus as a leading republican citizen , are all part of an effective and carefully constructed veneer . essay by dr. jeffrey a. becker additional resources : ara pacis museum ara pacis augustae ( reed college ) roma sparita photo archive ( italian ) ara pacis museum / richard meier & amp ; partners architects david castriota , the ara pacis augustae and the imagery of abundance in later greek and early roman imperial art ( princeton : princeton university press , 1995 ) . diane a. conlin , the artists of the ara pacis : the process of hellenization in roman relief sculpture ( chapel hill : university of north carolina press , 1997 ) .
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if so , how come the ara pacis `` meier pavilion has not been well-received '' , while the getty center is beloved ?
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question . she or he then comes up with a hypothesis , a testable explanation that addresses the question . a hypothesis is n't necessarily right . instead , it 's a `` best guess , '' and the scientist must test it to see if it 's actually correct . scientists test hypotheses by making predictions : if hypothesis $ \text x $ is right , then $ \text y $ should be true . then , they do experiments or make observations to see if the predictions are correct . if they are , the hypothesis is supported . if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments . a controlled experiment is a scientific test done under controlled conditions , meaning that just one ( or a few ) factors are changed at a time , while all others are kept constant . we 'll look closely at controlled experiments in the next section . in some cases , there is no good way to test a hypothesis using a controlled experiment ( for practical or ethical reasons ) . in that case , a scientist may test a hypothesis by making predictions about patterns that should be seen in nature if the hypothesis is correct . then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example . suppose i decide to grow bean sprouts in my kitchen , near the window . i put bean seeds in a pot with soil , set them on the windowsill , and wait for them to sprout . however , after several weeks , i have no sprouts . why not ? well ... it turns out i forgot to water the seeds . so , i hypothesize that they did n't sprout due to lack of water . to test my hypothesis , i do a controlled experiment . in this experiment , i set up two identical pots . both contain ten bean seeds planted in the same type of soil , and both are placed in the same window . in fact , there is only one thing that i do differently to the two pots : one pot of seeds gets watered every afternoon . the other pot of seeds does n't get any water at all . after a week , nine out of ten seeds in the watered pot have sprouted , while none of the seeds in the dry pot have sprouted . it looks like the `` seeds need water '' hypothesis is probably correct ! let 's see how this simple example illustrates the parts of a controlled experiment . control and experimental groups there are two groups in the experiment , and they are identical except that one receives a treatment ( water ) while the other does not . the group that receives the treatment in an experiment ( here , the watered pot ) is called the experimental group , while the group that does not receive the treatment ( here , the dry pot ) is called the control group . the control group provides a baseline that lets us see if the treatment has an effect . independent and dependent variables the factor that is different between the control and experimental groups ( in this case , the amount of water ) is known as the independent variable . this variable is independent because it does not depend on what happens in the experiment . instead , it is something that the experimenter applies or chooses him/herself . in contrast , the dependent variable in an experiment is the response that 's measured to see if the treatment had an effect . in this case , the fraction of bean seeds that sprouted is the dependent variable . the dependent variable ( fraction of seeds sprouting ) depends on the independent variable ( the amount of water ) , and not vice versa . experimental data ( singular : datum ) are observations made during the experiment . in this case , the data we collected were the number of bean sprouts in each pot after a week . variability and repetition out of the ten watered bean seeds , only nine came up . what happened to the tenth seed ? that seed may have been dead , unhealthy , or just slow to sprout . especially in biology ( which studies complex , living things ) , there is often variation in the material used for an experiment – here , the bean seeds – that the experimenter can not see . because of this potential for variation , biology experiments need to have a large sample size and , ideally , be repeated several times . sample size refers to the number of individual items tested in an experiment – in this case , $ 10 $ bean seeds per group . having more samples and repeating the experiment more times makes it less likely that we will reach a wrong conclusion because of random variation . biologists and other scientists also use statistical tests to help them distinguish real differences from differences due to random variation ( e.g. , when comparing experimental and control groups ) . controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ . however , a team of australian researchers hypothesized that other factors might be important too . specifically , they tested the hypothesis that high $ \text { co } _2 $ levels , which make ocean waters more acidic , might also promote bleaching $ ^2 $ . what kind of experiment would you do to test this hypothesis ? think about : what your control and experimental groups would be what your independent and dependent variables would be what results you would predict in each group have you given it a try ? non-experimental hypothesis tests some types of hypotheses ca n't be tested in controlled experiments for ethical or practical reasons . for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white . researchers suspected that high water temperature might cause bleaching , and they tested this hypothesis experimentally on a small scale ( using isolated coral fragments in tanks ) $ ^ { 3,4 } $ . what ecologists most wanted to know , however , was whether water temperature was causing bleaching for lots of different coral species in their natural setting . this broader question could not be answered experimentally , as it would n't be ethical ( or even possible ) to artificially change the water temperature surrounding entire coral reefs . instead , to test the hypothesis that natural bleaching events were caused by increases in water temperature , a team of researchers wrote a computer program to predict bleaching events based on real-time water temperature data . for example , this program would generally predict bleaching for a particular reef when the water temperature in the reef 's area exceeded its average monthly maximum by $ 1 $ $ ^\circ \text c $ or more $ ^ { 1 } $ . the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ .
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regarding the australian coral reef test of bleaching : how come the corals were 10 % bleached in normal seawater , should n't they have not been bleached at all considering it was normal acidity ?
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question . she or he then comes up with a hypothesis , a testable explanation that addresses the question . a hypothesis is n't necessarily right . instead , it 's a `` best guess , '' and the scientist must test it to see if it 's actually correct . scientists test hypotheses by making predictions : if hypothesis $ \text x $ is right , then $ \text y $ should be true . then , they do experiments or make observations to see if the predictions are correct . if they are , the hypothesis is supported . if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments . a controlled experiment is a scientific test done under controlled conditions , meaning that just one ( or a few ) factors are changed at a time , while all others are kept constant . we 'll look closely at controlled experiments in the next section . in some cases , there is no good way to test a hypothesis using a controlled experiment ( for practical or ethical reasons ) . in that case , a scientist may test a hypothesis by making predictions about patterns that should be seen in nature if the hypothesis is correct . then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example . suppose i decide to grow bean sprouts in my kitchen , near the window . i put bean seeds in a pot with soil , set them on the windowsill , and wait for them to sprout . however , after several weeks , i have no sprouts . why not ? well ... it turns out i forgot to water the seeds . so , i hypothesize that they did n't sprout due to lack of water . to test my hypothesis , i do a controlled experiment . in this experiment , i set up two identical pots . both contain ten bean seeds planted in the same type of soil , and both are placed in the same window . in fact , there is only one thing that i do differently to the two pots : one pot of seeds gets watered every afternoon . the other pot of seeds does n't get any water at all . after a week , nine out of ten seeds in the watered pot have sprouted , while none of the seeds in the dry pot have sprouted . it looks like the `` seeds need water '' hypothesis is probably correct ! let 's see how this simple example illustrates the parts of a controlled experiment . control and experimental groups there are two groups in the experiment , and they are identical except that one receives a treatment ( water ) while the other does not . the group that receives the treatment in an experiment ( here , the watered pot ) is called the experimental group , while the group that does not receive the treatment ( here , the dry pot ) is called the control group . the control group provides a baseline that lets us see if the treatment has an effect . independent and dependent variables the factor that is different between the control and experimental groups ( in this case , the amount of water ) is known as the independent variable . this variable is independent because it does not depend on what happens in the experiment . instead , it is something that the experimenter applies or chooses him/herself . in contrast , the dependent variable in an experiment is the response that 's measured to see if the treatment had an effect . in this case , the fraction of bean seeds that sprouted is the dependent variable . the dependent variable ( fraction of seeds sprouting ) depends on the independent variable ( the amount of water ) , and not vice versa . experimental data ( singular : datum ) are observations made during the experiment . in this case , the data we collected were the number of bean sprouts in each pot after a week . variability and repetition out of the ten watered bean seeds , only nine came up . what happened to the tenth seed ? that seed may have been dead , unhealthy , or just slow to sprout . especially in biology ( which studies complex , living things ) , there is often variation in the material used for an experiment – here , the bean seeds – that the experimenter can not see . because of this potential for variation , biology experiments need to have a large sample size and , ideally , be repeated several times . sample size refers to the number of individual items tested in an experiment – in this case , $ 10 $ bean seeds per group . having more samples and repeating the experiment more times makes it less likely that we will reach a wrong conclusion because of random variation . biologists and other scientists also use statistical tests to help them distinguish real differences from differences due to random variation ( e.g. , when comparing experimental and control groups ) . controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ . however , a team of australian researchers hypothesized that other factors might be important too . specifically , they tested the hypothesis that high $ \text { co } _2 $ levels , which make ocean waters more acidic , might also promote bleaching $ ^2 $ . what kind of experiment would you do to test this hypothesis ? think about : what your control and experimental groups would be what your independent and dependent variables would be what results you would predict in each group have you given it a try ? non-experimental hypothesis tests some types of hypotheses ca n't be tested in controlled experiments for ethical or practical reasons . for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white . researchers suspected that high water temperature might cause bleaching , and they tested this hypothesis experimentally on a small scale ( using isolated coral fragments in tanks ) $ ^ { 3,4 } $ . what ecologists most wanted to know , however , was whether water temperature was causing bleaching for lots of different coral species in their natural setting . this broader question could not be answered experimentally , as it would n't be ethical ( or even possible ) to artificially change the water temperature surrounding entire coral reefs . instead , to test the hypothesis that natural bleaching events were caused by increases in water temperature , a team of researchers wrote a computer program to predict bleaching events based on real-time water temperature data . for example , this program would generally predict bleaching for a particular reef when the water temperature in the reef 's area exceeded its average monthly maximum by $ 1 $ $ ^\circ \text c $ or more $ ^ { 1 } $ . the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present .
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can someone give an example of a non-experimental hypothesis test ?
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question . she or he then comes up with a hypothesis , a testable explanation that addresses the question . a hypothesis is n't necessarily right . instead , it 's a `` best guess , '' and the scientist must test it to see if it 's actually correct . scientists test hypotheses by making predictions : if hypothesis $ \text x $ is right , then $ \text y $ should be true . then , they do experiments or make observations to see if the predictions are correct . if they are , the hypothesis is supported . if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments . a controlled experiment is a scientific test done under controlled conditions , meaning that just one ( or a few ) factors are changed at a time , while all others are kept constant . we 'll look closely at controlled experiments in the next section . in some cases , there is no good way to test a hypothesis using a controlled experiment ( for practical or ethical reasons ) . in that case , a scientist may test a hypothesis by making predictions about patterns that should be seen in nature if the hypothesis is correct . then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example . suppose i decide to grow bean sprouts in my kitchen , near the window . i put bean seeds in a pot with soil , set them on the windowsill , and wait for them to sprout . however , after several weeks , i have no sprouts . why not ? well ... it turns out i forgot to water the seeds . so , i hypothesize that they did n't sprout due to lack of water . to test my hypothesis , i do a controlled experiment . in this experiment , i set up two identical pots . both contain ten bean seeds planted in the same type of soil , and both are placed in the same window . in fact , there is only one thing that i do differently to the two pots : one pot of seeds gets watered every afternoon . the other pot of seeds does n't get any water at all . after a week , nine out of ten seeds in the watered pot have sprouted , while none of the seeds in the dry pot have sprouted . it looks like the `` seeds need water '' hypothesis is probably correct ! let 's see how this simple example illustrates the parts of a controlled experiment . control and experimental groups there are two groups in the experiment , and they are identical except that one receives a treatment ( water ) while the other does not . the group that receives the treatment in an experiment ( here , the watered pot ) is called the experimental group , while the group that does not receive the treatment ( here , the dry pot ) is called the control group . the control group provides a baseline that lets us see if the treatment has an effect . independent and dependent variables the factor that is different between the control and experimental groups ( in this case , the amount of water ) is known as the independent variable . this variable is independent because it does not depend on what happens in the experiment . instead , it is something that the experimenter applies or chooses him/herself . in contrast , the dependent variable in an experiment is the response that 's measured to see if the treatment had an effect . in this case , the fraction of bean seeds that sprouted is the dependent variable . the dependent variable ( fraction of seeds sprouting ) depends on the independent variable ( the amount of water ) , and not vice versa . experimental data ( singular : datum ) are observations made during the experiment . in this case , the data we collected were the number of bean sprouts in each pot after a week . variability and repetition out of the ten watered bean seeds , only nine came up . what happened to the tenth seed ? that seed may have been dead , unhealthy , or just slow to sprout . especially in biology ( which studies complex , living things ) , there is often variation in the material used for an experiment – here , the bean seeds – that the experimenter can not see . because of this potential for variation , biology experiments need to have a large sample size and , ideally , be repeated several times . sample size refers to the number of individual items tested in an experiment – in this case , $ 10 $ bean seeds per group . having more samples and repeating the experiment more times makes it less likely that we will reach a wrong conclusion because of random variation . biologists and other scientists also use statistical tests to help them distinguish real differences from differences due to random variation ( e.g. , when comparing experimental and control groups ) . controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ . however , a team of australian researchers hypothesized that other factors might be important too . specifically , they tested the hypothesis that high $ \text { co } _2 $ levels , which make ocean waters more acidic , might also promote bleaching $ ^2 $ . what kind of experiment would you do to test this hypothesis ? think about : what your control and experimental groups would be what your independent and dependent variables would be what results you would predict in each group have you given it a try ? non-experimental hypothesis tests some types of hypotheses ca n't be tested in controlled experiments for ethical or practical reasons . for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white . researchers suspected that high water temperature might cause bleaching , and they tested this hypothesis experimentally on a small scale ( using isolated coral fragments in tanks ) $ ^ { 3,4 } $ . what ecologists most wanted to know , however , was whether water temperature was causing bleaching for lots of different coral species in their natural setting . this broader question could not be answered experimentally , as it would n't be ethical ( or even possible ) to artificially change the water temperature surrounding entire coral reefs . instead , to test the hypothesis that natural bleaching events were caused by increases in water temperature , a team of researchers wrote a computer program to predict bleaching events based on real-time water temperature data . for example , this program would generally predict bleaching for a particular reef when the water temperature in the reef 's area exceeded its average monthly maximum by $ 1 $ $ ^\circ \text c $ or more $ ^ { 1 } $ . the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments .
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so hypotheses are like conditional statements ?
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question . she or he then comes up with a hypothesis , a testable explanation that addresses the question . a hypothesis is n't necessarily right . instead , it 's a `` best guess , '' and the scientist must test it to see if it 's actually correct . scientists test hypotheses by making predictions : if hypothesis $ \text x $ is right , then $ \text y $ should be true . then , they do experiments or make observations to see if the predictions are correct . if they are , the hypothesis is supported . if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments . a controlled experiment is a scientific test done under controlled conditions , meaning that just one ( or a few ) factors are changed at a time , while all others are kept constant . we 'll look closely at controlled experiments in the next section . in some cases , there is no good way to test a hypothesis using a controlled experiment ( for practical or ethical reasons ) . in that case , a scientist may test a hypothesis by making predictions about patterns that should be seen in nature if the hypothesis is correct . then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example . suppose i decide to grow bean sprouts in my kitchen , near the window . i put bean seeds in a pot with soil , set them on the windowsill , and wait for them to sprout . however , after several weeks , i have no sprouts . why not ? well ... it turns out i forgot to water the seeds . so , i hypothesize that they did n't sprout due to lack of water . to test my hypothesis , i do a controlled experiment . in this experiment , i set up two identical pots . both contain ten bean seeds planted in the same type of soil , and both are placed in the same window . in fact , there is only one thing that i do differently to the two pots : one pot of seeds gets watered every afternoon . the other pot of seeds does n't get any water at all . after a week , nine out of ten seeds in the watered pot have sprouted , while none of the seeds in the dry pot have sprouted . it looks like the `` seeds need water '' hypothesis is probably correct ! let 's see how this simple example illustrates the parts of a controlled experiment . control and experimental groups there are two groups in the experiment , and they are identical except that one receives a treatment ( water ) while the other does not . the group that receives the treatment in an experiment ( here , the watered pot ) is called the experimental group , while the group that does not receive the treatment ( here , the dry pot ) is called the control group . the control group provides a baseline that lets us see if the treatment has an effect . independent and dependent variables the factor that is different between the control and experimental groups ( in this case , the amount of water ) is known as the independent variable . this variable is independent because it does not depend on what happens in the experiment . instead , it is something that the experimenter applies or chooses him/herself . in contrast , the dependent variable in an experiment is the response that 's measured to see if the treatment had an effect . in this case , the fraction of bean seeds that sprouted is the dependent variable . the dependent variable ( fraction of seeds sprouting ) depends on the independent variable ( the amount of water ) , and not vice versa . experimental data ( singular : datum ) are observations made during the experiment . in this case , the data we collected were the number of bean sprouts in each pot after a week . variability and repetition out of the ten watered bean seeds , only nine came up . what happened to the tenth seed ? that seed may have been dead , unhealthy , or just slow to sprout . especially in biology ( which studies complex , living things ) , there is often variation in the material used for an experiment – here , the bean seeds – that the experimenter can not see . because of this potential for variation , biology experiments need to have a large sample size and , ideally , be repeated several times . sample size refers to the number of individual items tested in an experiment – in this case , $ 10 $ bean seeds per group . having more samples and repeating the experiment more times makes it less likely that we will reach a wrong conclusion because of random variation . biologists and other scientists also use statistical tests to help them distinguish real differences from differences due to random variation ( e.g. , when comparing experimental and control groups ) . controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ . however , a team of australian researchers hypothesized that other factors might be important too . specifically , they tested the hypothesis that high $ \text { co } _2 $ levels , which make ocean waters more acidic , might also promote bleaching $ ^2 $ . what kind of experiment would you do to test this hypothesis ? think about : what your control and experimental groups would be what your independent and dependent variables would be what results you would predict in each group have you given it a try ? non-experimental hypothesis tests some types of hypotheses ca n't be tested in controlled experiments for ethical or practical reasons . for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white . researchers suspected that high water temperature might cause bleaching , and they tested this hypothesis experimentally on a small scale ( using isolated coral fragments in tanks ) $ ^ { 3,4 } $ . what ecologists most wanted to know , however , was whether water temperature was causing bleaching for lots of different coral species in their natural setting . this broader question could not be answered experimentally , as it would n't be ethical ( or even possible ) to artificially change the water temperature surrounding entire coral reefs . instead , to test the hypothesis that natural bleaching events were caused by increases in water temperature , a team of researchers wrote a computer program to predict bleaching events based on real-time water temperature data . for example , this program would generally predict bleaching for a particular reef when the water temperature in the reef 's area exceeded its average monthly maximum by $ 1 $ $ ^\circ \text c $ or more $ ^ { 1 } $ . the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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is is possible that both hypotheses are correct , and that the co2 effect is actually a result of the temperature-induced bleaching ?
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question . she or he then comes up with a hypothesis , a testable explanation that addresses the question . a hypothesis is n't necessarily right . instead , it 's a `` best guess , '' and the scientist must test it to see if it 's actually correct . scientists test hypotheses by making predictions : if hypothesis $ \text x $ is right , then $ \text y $ should be true . then , they do experiments or make observations to see if the predictions are correct . if they are , the hypothesis is supported . if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments . a controlled experiment is a scientific test done under controlled conditions , meaning that just one ( or a few ) factors are changed at a time , while all others are kept constant . we 'll look closely at controlled experiments in the next section . in some cases , there is no good way to test a hypothesis using a controlled experiment ( for practical or ethical reasons ) . in that case , a scientist may test a hypothesis by making predictions about patterns that should be seen in nature if the hypothesis is correct . then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example . suppose i decide to grow bean sprouts in my kitchen , near the window . i put bean seeds in a pot with soil , set them on the windowsill , and wait for them to sprout . however , after several weeks , i have no sprouts . why not ? well ... it turns out i forgot to water the seeds . so , i hypothesize that they did n't sprout due to lack of water . to test my hypothesis , i do a controlled experiment . in this experiment , i set up two identical pots . both contain ten bean seeds planted in the same type of soil , and both are placed in the same window . in fact , there is only one thing that i do differently to the two pots : one pot of seeds gets watered every afternoon . the other pot of seeds does n't get any water at all . after a week , nine out of ten seeds in the watered pot have sprouted , while none of the seeds in the dry pot have sprouted . it looks like the `` seeds need water '' hypothesis is probably correct ! let 's see how this simple example illustrates the parts of a controlled experiment . control and experimental groups there are two groups in the experiment , and they are identical except that one receives a treatment ( water ) while the other does not . the group that receives the treatment in an experiment ( here , the watered pot ) is called the experimental group , while the group that does not receive the treatment ( here , the dry pot ) is called the control group . the control group provides a baseline that lets us see if the treatment has an effect . independent and dependent variables the factor that is different between the control and experimental groups ( in this case , the amount of water ) is known as the independent variable . this variable is independent because it does not depend on what happens in the experiment . instead , it is something that the experimenter applies or chooses him/herself . in contrast , the dependent variable in an experiment is the response that 's measured to see if the treatment had an effect . in this case , the fraction of bean seeds that sprouted is the dependent variable . the dependent variable ( fraction of seeds sprouting ) depends on the independent variable ( the amount of water ) , and not vice versa . experimental data ( singular : datum ) are observations made during the experiment . in this case , the data we collected were the number of bean sprouts in each pot after a week . variability and repetition out of the ten watered bean seeds , only nine came up . what happened to the tenth seed ? that seed may have been dead , unhealthy , or just slow to sprout . especially in biology ( which studies complex , living things ) , there is often variation in the material used for an experiment – here , the bean seeds – that the experimenter can not see . because of this potential for variation , biology experiments need to have a large sample size and , ideally , be repeated several times . sample size refers to the number of individual items tested in an experiment – in this case , $ 10 $ bean seeds per group . having more samples and repeating the experiment more times makes it less likely that we will reach a wrong conclusion because of random variation . biologists and other scientists also use statistical tests to help them distinguish real differences from differences due to random variation ( e.g. , when comparing experimental and control groups ) . controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ . however , a team of australian researchers hypothesized that other factors might be important too . specifically , they tested the hypothesis that high $ \text { co } _2 $ levels , which make ocean waters more acidic , might also promote bleaching $ ^2 $ . what kind of experiment would you do to test this hypothesis ? think about : what your control and experimental groups would be what your independent and dependent variables would be what results you would predict in each group have you given it a try ? non-experimental hypothesis tests some types of hypotheses ca n't be tested in controlled experiments for ethical or practical reasons . for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white . researchers suspected that high water temperature might cause bleaching , and they tested this hypothesis experimentally on a small scale ( using isolated coral fragments in tanks ) $ ^ { 3,4 } $ . what ecologists most wanted to know , however , was whether water temperature was causing bleaching for lots of different coral species in their natural setting . this broader question could not be answered experimentally , as it would n't be ethical ( or even possible ) to artificially change the water temperature surrounding entire coral reefs . instead , to test the hypothesis that natural bleaching events were caused by increases in water temperature , a team of researchers wrote a computer program to predict bleaching events based on real-time water temperature data . for example , this program would generally predict bleaching for a particular reef when the water temperature in the reef 's area exceeded its average monthly maximum by $ 1 $ $ ^\circ \text c $ or more $ ^ { 1 } $ . the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example .
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what are the key ingredients of a controlled experiment ?
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question . she or he then comes up with a hypothesis , a testable explanation that addresses the question . a hypothesis is n't necessarily right . instead , it 's a `` best guess , '' and the scientist must test it to see if it 's actually correct . scientists test hypotheses by making predictions : if hypothesis $ \text x $ is right , then $ \text y $ should be true . then , they do experiments or make observations to see if the predictions are correct . if they are , the hypothesis is supported . if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments . a controlled experiment is a scientific test done under controlled conditions , meaning that just one ( or a few ) factors are changed at a time , while all others are kept constant . we 'll look closely at controlled experiments in the next section . in some cases , there is no good way to test a hypothesis using a controlled experiment ( for practical or ethical reasons ) . in that case , a scientist may test a hypothesis by making predictions about patterns that should be seen in nature if the hypothesis is correct . then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example . suppose i decide to grow bean sprouts in my kitchen , near the window . i put bean seeds in a pot with soil , set them on the windowsill , and wait for them to sprout . however , after several weeks , i have no sprouts . why not ? well ... it turns out i forgot to water the seeds . so , i hypothesize that they did n't sprout due to lack of water . to test my hypothesis , i do a controlled experiment . in this experiment , i set up two identical pots . both contain ten bean seeds planted in the same type of soil , and both are placed in the same window . in fact , there is only one thing that i do differently to the two pots : one pot of seeds gets watered every afternoon . the other pot of seeds does n't get any water at all . after a week , nine out of ten seeds in the watered pot have sprouted , while none of the seeds in the dry pot have sprouted . it looks like the `` seeds need water '' hypothesis is probably correct ! let 's see how this simple example illustrates the parts of a controlled experiment . control and experimental groups there are two groups in the experiment , and they are identical except that one receives a treatment ( water ) while the other does not . the group that receives the treatment in an experiment ( here , the watered pot ) is called the experimental group , while the group that does not receive the treatment ( here , the dry pot ) is called the control group . the control group provides a baseline that lets us see if the treatment has an effect . independent and dependent variables the factor that is different between the control and experimental groups ( in this case , the amount of water ) is known as the independent variable . this variable is independent because it does not depend on what happens in the experiment . instead , it is something that the experimenter applies or chooses him/herself . in contrast , the dependent variable in an experiment is the response that 's measured to see if the treatment had an effect . in this case , the fraction of bean seeds that sprouted is the dependent variable . the dependent variable ( fraction of seeds sprouting ) depends on the independent variable ( the amount of water ) , and not vice versa . experimental data ( singular : datum ) are observations made during the experiment . in this case , the data we collected were the number of bean sprouts in each pot after a week . variability and repetition out of the ten watered bean seeds , only nine came up . what happened to the tenth seed ? that seed may have been dead , unhealthy , or just slow to sprout . especially in biology ( which studies complex , living things ) , there is often variation in the material used for an experiment – here , the bean seeds – that the experimenter can not see . because of this potential for variation , biology experiments need to have a large sample size and , ideally , be repeated several times . sample size refers to the number of individual items tested in an experiment – in this case , $ 10 $ bean seeds per group . having more samples and repeating the experiment more times makes it less likely that we will reach a wrong conclusion because of random variation . biologists and other scientists also use statistical tests to help them distinguish real differences from differences due to random variation ( e.g. , when comparing experimental and control groups ) . controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ . however , a team of australian researchers hypothesized that other factors might be important too . specifically , they tested the hypothesis that high $ \text { co } _2 $ levels , which make ocean waters more acidic , might also promote bleaching $ ^2 $ . what kind of experiment would you do to test this hypothesis ? think about : what your control and experimental groups would be what your independent and dependent variables would be what results you would predict in each group have you given it a try ? non-experimental hypothesis tests some types of hypotheses ca n't be tested in controlled experiments for ethical or practical reasons . for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white . researchers suspected that high water temperature might cause bleaching , and they tested this hypothesis experimentally on a small scale ( using isolated coral fragments in tanks ) $ ^ { 3,4 } $ . what ecologists most wanted to know , however , was whether water temperature was causing bleaching for lots of different coral species in their natural setting . this broader question could not be answered experimentally , as it would n't be ethical ( or even possible ) to artificially change the water temperature surrounding entire coral reefs . instead , to test the hypothesis that natural bleaching events were caused by increases in water temperature , a team of researchers wrote a computer program to predict bleaching events based on real-time water temperature data . for example , this program would generally predict bleaching for a particular reef when the water temperature in the reef 's area exceeded its average monthly maximum by $ 1 $ $ ^\circ \text c $ or more $ ^ { 1 } $ . the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question .
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what is the organism in the picture ?
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question . she or he then comes up with a hypothesis , a testable explanation that addresses the question . a hypothesis is n't necessarily right . instead , it 's a `` best guess , '' and the scientist must test it to see if it 's actually correct . scientists test hypotheses by making predictions : if hypothesis $ \text x $ is right , then $ \text y $ should be true . then , they do experiments or make observations to see if the predictions are correct . if they are , the hypothesis is supported . if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments . a controlled experiment is a scientific test done under controlled conditions , meaning that just one ( or a few ) factors are changed at a time , while all others are kept constant . we 'll look closely at controlled experiments in the next section . in some cases , there is no good way to test a hypothesis using a controlled experiment ( for practical or ethical reasons ) . in that case , a scientist may test a hypothesis by making predictions about patterns that should be seen in nature if the hypothesis is correct . then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example . suppose i decide to grow bean sprouts in my kitchen , near the window . i put bean seeds in a pot with soil , set them on the windowsill , and wait for them to sprout . however , after several weeks , i have no sprouts . why not ? well ... it turns out i forgot to water the seeds . so , i hypothesize that they did n't sprout due to lack of water . to test my hypothesis , i do a controlled experiment . in this experiment , i set up two identical pots . both contain ten bean seeds planted in the same type of soil , and both are placed in the same window . in fact , there is only one thing that i do differently to the two pots : one pot of seeds gets watered every afternoon . the other pot of seeds does n't get any water at all . after a week , nine out of ten seeds in the watered pot have sprouted , while none of the seeds in the dry pot have sprouted . it looks like the `` seeds need water '' hypothesis is probably correct ! let 's see how this simple example illustrates the parts of a controlled experiment . control and experimental groups there are two groups in the experiment , and they are identical except that one receives a treatment ( water ) while the other does not . the group that receives the treatment in an experiment ( here , the watered pot ) is called the experimental group , while the group that does not receive the treatment ( here , the dry pot ) is called the control group . the control group provides a baseline that lets us see if the treatment has an effect . independent and dependent variables the factor that is different between the control and experimental groups ( in this case , the amount of water ) is known as the independent variable . this variable is independent because it does not depend on what happens in the experiment . instead , it is something that the experimenter applies or chooses him/herself . in contrast , the dependent variable in an experiment is the response that 's measured to see if the treatment had an effect . in this case , the fraction of bean seeds that sprouted is the dependent variable . the dependent variable ( fraction of seeds sprouting ) depends on the independent variable ( the amount of water ) , and not vice versa . experimental data ( singular : datum ) are observations made during the experiment . in this case , the data we collected were the number of bean sprouts in each pot after a week . variability and repetition out of the ten watered bean seeds , only nine came up . what happened to the tenth seed ? that seed may have been dead , unhealthy , or just slow to sprout . especially in biology ( which studies complex , living things ) , there is often variation in the material used for an experiment – here , the bean seeds – that the experimenter can not see . because of this potential for variation , biology experiments need to have a large sample size and , ideally , be repeated several times . sample size refers to the number of individual items tested in an experiment – in this case , $ 10 $ bean seeds per group . having more samples and repeating the experiment more times makes it less likely that we will reach a wrong conclusion because of random variation . biologists and other scientists also use statistical tests to help them distinguish real differences from differences due to random variation ( e.g. , when comparing experimental and control groups ) . controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ . however , a team of australian researchers hypothesized that other factors might be important too . specifically , they tested the hypothesis that high $ \text { co } _2 $ levels , which make ocean waters more acidic , might also promote bleaching $ ^2 $ . what kind of experiment would you do to test this hypothesis ? think about : what your control and experimental groups would be what your independent and dependent variables would be what results you would predict in each group have you given it a try ? non-experimental hypothesis tests some types of hypotheses ca n't be tested in controlled experiments for ethical or practical reasons . for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white . researchers suspected that high water temperature might cause bleaching , and they tested this hypothesis experimentally on a small scale ( using isolated coral fragments in tanks ) $ ^ { 3,4 } $ . what ecologists most wanted to know , however , was whether water temperature was causing bleaching for lots of different coral species in their natural setting . this broader question could not be answered experimentally , as it would n't be ethical ( or even possible ) to artificially change the water temperature surrounding entire coral reefs . instead , to test the hypothesis that natural bleaching events were caused by increases in water temperature , a team of researchers wrote a computer program to predict bleaching events based on real-time water temperature data . for example , this program would generally predict bleaching for a particular reef when the water temperature in the reef 's area exceeded its average monthly maximum by $ 1 $ $ ^\circ \text c $ or more $ ^ { 1 } $ . the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ .
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when the coral reefs are bleached does that mean they have died , if not do they stay white forever or return to their original form ?
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question . she or he then comes up with a hypothesis , a testable explanation that addresses the question . a hypothesis is n't necessarily right . instead , it 's a `` best guess , '' and the scientist must test it to see if it 's actually correct . scientists test hypotheses by making predictions : if hypothesis $ \text x $ is right , then $ \text y $ should be true . then , they do experiments or make observations to see if the predictions are correct . if they are , the hypothesis is supported . if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments . a controlled experiment is a scientific test done under controlled conditions , meaning that just one ( or a few ) factors are changed at a time , while all others are kept constant . we 'll look closely at controlled experiments in the next section . in some cases , there is no good way to test a hypothesis using a controlled experiment ( for practical or ethical reasons ) . in that case , a scientist may test a hypothesis by making predictions about patterns that should be seen in nature if the hypothesis is correct . then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example . suppose i decide to grow bean sprouts in my kitchen , near the window . i put bean seeds in a pot with soil , set them on the windowsill , and wait for them to sprout . however , after several weeks , i have no sprouts . why not ? well ... it turns out i forgot to water the seeds . so , i hypothesize that they did n't sprout due to lack of water . to test my hypothesis , i do a controlled experiment . in this experiment , i set up two identical pots . both contain ten bean seeds planted in the same type of soil , and both are placed in the same window . in fact , there is only one thing that i do differently to the two pots : one pot of seeds gets watered every afternoon . the other pot of seeds does n't get any water at all . after a week , nine out of ten seeds in the watered pot have sprouted , while none of the seeds in the dry pot have sprouted . it looks like the `` seeds need water '' hypothesis is probably correct ! let 's see how this simple example illustrates the parts of a controlled experiment . control and experimental groups there are two groups in the experiment , and they are identical except that one receives a treatment ( water ) while the other does not . the group that receives the treatment in an experiment ( here , the watered pot ) is called the experimental group , while the group that does not receive the treatment ( here , the dry pot ) is called the control group . the control group provides a baseline that lets us see if the treatment has an effect . independent and dependent variables the factor that is different between the control and experimental groups ( in this case , the amount of water ) is known as the independent variable . this variable is independent because it does not depend on what happens in the experiment . instead , it is something that the experimenter applies or chooses him/herself . in contrast , the dependent variable in an experiment is the response that 's measured to see if the treatment had an effect . in this case , the fraction of bean seeds that sprouted is the dependent variable . the dependent variable ( fraction of seeds sprouting ) depends on the independent variable ( the amount of water ) , and not vice versa . experimental data ( singular : datum ) are observations made during the experiment . in this case , the data we collected were the number of bean sprouts in each pot after a week . variability and repetition out of the ten watered bean seeds , only nine came up . what happened to the tenth seed ? that seed may have been dead , unhealthy , or just slow to sprout . especially in biology ( which studies complex , living things ) , there is often variation in the material used for an experiment – here , the bean seeds – that the experimenter can not see . because of this potential for variation , biology experiments need to have a large sample size and , ideally , be repeated several times . sample size refers to the number of individual items tested in an experiment – in this case , $ 10 $ bean seeds per group . having more samples and repeating the experiment more times makes it less likely that we will reach a wrong conclusion because of random variation . biologists and other scientists also use statistical tests to help them distinguish real differences from differences due to random variation ( e.g. , when comparing experimental and control groups ) . controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ . however , a team of australian researchers hypothesized that other factors might be important too . specifically , they tested the hypothesis that high $ \text { co } _2 $ levels , which make ocean waters more acidic , might also promote bleaching $ ^2 $ . what kind of experiment would you do to test this hypothesis ? think about : what your control and experimental groups would be what your independent and dependent variables would be what results you would predict in each group have you given it a try ? non-experimental hypothesis tests some types of hypotheses ca n't be tested in controlled experiments for ethical or practical reasons . for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white . researchers suspected that high water temperature might cause bleaching , and they tested this hypothesis experimentally on a small scale ( using isolated coral fragments in tanks ) $ ^ { 3,4 } $ . what ecologists most wanted to know , however , was whether water temperature was causing bleaching for lots of different coral species in their natural setting . this broader question could not be answered experimentally , as it would n't be ethical ( or even possible ) to artificially change the water temperature surrounding entire coral reefs . instead , to test the hypothesis that natural bleaching events were caused by increases in water temperature , a team of researchers wrote a computer program to predict bleaching events based on real-time water temperature data . for example , this program would generally predict bleaching for a particular reef when the water temperature in the reef 's area exceeded its average monthly maximum by $ 1 $ $ ^\circ \text c $ or more $ ^ { 1 } $ . the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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what 's the difference between bleaching and temperature ?
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question . she or he then comes up with a hypothesis , a testable explanation that addresses the question . a hypothesis is n't necessarily right . instead , it 's a `` best guess , '' and the scientist must test it to see if it 's actually correct . scientists test hypotheses by making predictions : if hypothesis $ \text x $ is right , then $ \text y $ should be true . then , they do experiments or make observations to see if the predictions are correct . if they are , the hypothesis is supported . if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments . a controlled experiment is a scientific test done under controlled conditions , meaning that just one ( or a few ) factors are changed at a time , while all others are kept constant . we 'll look closely at controlled experiments in the next section . in some cases , there is no good way to test a hypothesis using a controlled experiment ( for practical or ethical reasons ) . in that case , a scientist may test a hypothesis by making predictions about patterns that should be seen in nature if the hypothesis is correct . then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example . suppose i decide to grow bean sprouts in my kitchen , near the window . i put bean seeds in a pot with soil , set them on the windowsill , and wait for them to sprout . however , after several weeks , i have no sprouts . why not ? well ... it turns out i forgot to water the seeds . so , i hypothesize that they did n't sprout due to lack of water . to test my hypothesis , i do a controlled experiment . in this experiment , i set up two identical pots . both contain ten bean seeds planted in the same type of soil , and both are placed in the same window . in fact , there is only one thing that i do differently to the two pots : one pot of seeds gets watered every afternoon . the other pot of seeds does n't get any water at all . after a week , nine out of ten seeds in the watered pot have sprouted , while none of the seeds in the dry pot have sprouted . it looks like the `` seeds need water '' hypothesis is probably correct ! let 's see how this simple example illustrates the parts of a controlled experiment . control and experimental groups there are two groups in the experiment , and they are identical except that one receives a treatment ( water ) while the other does not . the group that receives the treatment in an experiment ( here , the watered pot ) is called the experimental group , while the group that does not receive the treatment ( here , the dry pot ) is called the control group . the control group provides a baseline that lets us see if the treatment has an effect . independent and dependent variables the factor that is different between the control and experimental groups ( in this case , the amount of water ) is known as the independent variable . this variable is independent because it does not depend on what happens in the experiment . instead , it is something that the experimenter applies or chooses him/herself . in contrast , the dependent variable in an experiment is the response that 's measured to see if the treatment had an effect . in this case , the fraction of bean seeds that sprouted is the dependent variable . the dependent variable ( fraction of seeds sprouting ) depends on the independent variable ( the amount of water ) , and not vice versa . experimental data ( singular : datum ) are observations made during the experiment . in this case , the data we collected were the number of bean sprouts in each pot after a week . variability and repetition out of the ten watered bean seeds , only nine came up . what happened to the tenth seed ? that seed may have been dead , unhealthy , or just slow to sprout . especially in biology ( which studies complex , living things ) , there is often variation in the material used for an experiment – here , the bean seeds – that the experimenter can not see . because of this potential for variation , biology experiments need to have a large sample size and , ideally , be repeated several times . sample size refers to the number of individual items tested in an experiment – in this case , $ 10 $ bean seeds per group . having more samples and repeating the experiment more times makes it less likely that we will reach a wrong conclusion because of random variation . biologists and other scientists also use statistical tests to help them distinguish real differences from differences due to random variation ( e.g. , when comparing experimental and control groups ) . controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ . however , a team of australian researchers hypothesized that other factors might be important too . specifically , they tested the hypothesis that high $ \text { co } _2 $ levels , which make ocean waters more acidic , might also promote bleaching $ ^2 $ . what kind of experiment would you do to test this hypothesis ? think about : what your control and experimental groups would be what your independent and dependent variables would be what results you would predict in each group have you given it a try ? non-experimental hypothesis tests some types of hypotheses ca n't be tested in controlled experiments for ethical or practical reasons . for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white . researchers suspected that high water temperature might cause bleaching , and they tested this hypothesis experimentally on a small scale ( using isolated coral fragments in tanks ) $ ^ { 3,4 } $ . what ecologists most wanted to know , however , was whether water temperature was causing bleaching for lots of different coral species in their natural setting . this broader question could not be answered experimentally , as it would n't be ethical ( or even possible ) to artificially change the water temperature surrounding entire coral reefs . instead , to test the hypothesis that natural bleaching events were caused by increases in water temperature , a team of researchers wrote a computer program to predict bleaching events based on real-time water temperature data . for example , this program would generally predict bleaching for a particular reef when the water temperature in the reef 's area exceeded its average monthly maximum by $ 1 $ $ ^\circ \text c $ or more $ ^ { 1 } $ . the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing .
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so , why would one part of a coral reef bleach while mere inches away another part did n't ?
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question . she or he then comes up with a hypothesis , a testable explanation that addresses the question . a hypothesis is n't necessarily right . instead , it 's a `` best guess , '' and the scientist must test it to see if it 's actually correct . scientists test hypotheses by making predictions : if hypothesis $ \text x $ is right , then $ \text y $ should be true . then , they do experiments or make observations to see if the predictions are correct . if they are , the hypothesis is supported . if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments . a controlled experiment is a scientific test done under controlled conditions , meaning that just one ( or a few ) factors are changed at a time , while all others are kept constant . we 'll look closely at controlled experiments in the next section . in some cases , there is no good way to test a hypothesis using a controlled experiment ( for practical or ethical reasons ) . in that case , a scientist may test a hypothesis by making predictions about patterns that should be seen in nature if the hypothesis is correct . then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example . suppose i decide to grow bean sprouts in my kitchen , near the window . i put bean seeds in a pot with soil , set them on the windowsill , and wait for them to sprout . however , after several weeks , i have no sprouts . why not ? well ... it turns out i forgot to water the seeds . so , i hypothesize that they did n't sprout due to lack of water . to test my hypothesis , i do a controlled experiment . in this experiment , i set up two identical pots . both contain ten bean seeds planted in the same type of soil , and both are placed in the same window . in fact , there is only one thing that i do differently to the two pots : one pot of seeds gets watered every afternoon . the other pot of seeds does n't get any water at all . after a week , nine out of ten seeds in the watered pot have sprouted , while none of the seeds in the dry pot have sprouted . it looks like the `` seeds need water '' hypothesis is probably correct ! let 's see how this simple example illustrates the parts of a controlled experiment . control and experimental groups there are two groups in the experiment , and they are identical except that one receives a treatment ( water ) while the other does not . the group that receives the treatment in an experiment ( here , the watered pot ) is called the experimental group , while the group that does not receive the treatment ( here , the dry pot ) is called the control group . the control group provides a baseline that lets us see if the treatment has an effect . independent and dependent variables the factor that is different between the control and experimental groups ( in this case , the amount of water ) is known as the independent variable . this variable is independent because it does not depend on what happens in the experiment . instead , it is something that the experimenter applies or chooses him/herself . in contrast , the dependent variable in an experiment is the response that 's measured to see if the treatment had an effect . in this case , the fraction of bean seeds that sprouted is the dependent variable . the dependent variable ( fraction of seeds sprouting ) depends on the independent variable ( the amount of water ) , and not vice versa . experimental data ( singular : datum ) are observations made during the experiment . in this case , the data we collected were the number of bean sprouts in each pot after a week . variability and repetition out of the ten watered bean seeds , only nine came up . what happened to the tenth seed ? that seed may have been dead , unhealthy , or just slow to sprout . especially in biology ( which studies complex , living things ) , there is often variation in the material used for an experiment – here , the bean seeds – that the experimenter can not see . because of this potential for variation , biology experiments need to have a large sample size and , ideally , be repeated several times . sample size refers to the number of individual items tested in an experiment – in this case , $ 10 $ bean seeds per group . having more samples and repeating the experiment more times makes it less likely that we will reach a wrong conclusion because of random variation . biologists and other scientists also use statistical tests to help them distinguish real differences from differences due to random variation ( e.g. , when comparing experimental and control groups ) . controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ . however , a team of australian researchers hypothesized that other factors might be important too . specifically , they tested the hypothesis that high $ \text { co } _2 $ levels , which make ocean waters more acidic , might also promote bleaching $ ^2 $ . what kind of experiment would you do to test this hypothesis ? think about : what your control and experimental groups would be what your independent and dependent variables would be what results you would predict in each group have you given it a try ? non-experimental hypothesis tests some types of hypotheses ca n't be tested in controlled experiments for ethical or practical reasons . for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white . researchers suspected that high water temperature might cause bleaching , and they tested this hypothesis experimentally on a small scale ( using isolated coral fragments in tanks ) $ ^ { 3,4 } $ . what ecologists most wanted to know , however , was whether water temperature was causing bleaching for lots of different coral species in their natural setting . this broader question could not be answered experimentally , as it would n't be ethical ( or even possible ) to artificially change the water temperature surrounding entire coral reefs . instead , to test the hypothesis that natural bleaching events were caused by increases in water temperature , a team of researchers wrote a computer program to predict bleaching events based on real-time water temperature data . for example , this program would generally predict bleaching for a particular reef when the water temperature in the reef 's area exceeded its average monthly maximum by $ 1 $ $ ^\circ \text c $ or more $ ^ { 1 } $ . the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question .
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is the scientific method like a problem solving method to solve questions about the natural world ?
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question . she or he then comes up with a hypothesis , a testable explanation that addresses the question . a hypothesis is n't necessarily right . instead , it 's a `` best guess , '' and the scientist must test it to see if it 's actually correct . scientists test hypotheses by making predictions : if hypothesis $ \text x $ is right , then $ \text y $ should be true . then , they do experiments or make observations to see if the predictions are correct . if they are , the hypothesis is supported . if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments . a controlled experiment is a scientific test done under controlled conditions , meaning that just one ( or a few ) factors are changed at a time , while all others are kept constant . we 'll look closely at controlled experiments in the next section . in some cases , there is no good way to test a hypothesis using a controlled experiment ( for practical or ethical reasons ) . in that case , a scientist may test a hypothesis by making predictions about patterns that should be seen in nature if the hypothesis is correct . then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example . suppose i decide to grow bean sprouts in my kitchen , near the window . i put bean seeds in a pot with soil , set them on the windowsill , and wait for them to sprout . however , after several weeks , i have no sprouts . why not ? well ... it turns out i forgot to water the seeds . so , i hypothesize that they did n't sprout due to lack of water . to test my hypothesis , i do a controlled experiment . in this experiment , i set up two identical pots . both contain ten bean seeds planted in the same type of soil , and both are placed in the same window . in fact , there is only one thing that i do differently to the two pots : one pot of seeds gets watered every afternoon . the other pot of seeds does n't get any water at all . after a week , nine out of ten seeds in the watered pot have sprouted , while none of the seeds in the dry pot have sprouted . it looks like the `` seeds need water '' hypothesis is probably correct ! let 's see how this simple example illustrates the parts of a controlled experiment . control and experimental groups there are two groups in the experiment , and they are identical except that one receives a treatment ( water ) while the other does not . the group that receives the treatment in an experiment ( here , the watered pot ) is called the experimental group , while the group that does not receive the treatment ( here , the dry pot ) is called the control group . the control group provides a baseline that lets us see if the treatment has an effect . independent and dependent variables the factor that is different between the control and experimental groups ( in this case , the amount of water ) is known as the independent variable . this variable is independent because it does not depend on what happens in the experiment . instead , it is something that the experimenter applies or chooses him/herself . in contrast , the dependent variable in an experiment is the response that 's measured to see if the treatment had an effect . in this case , the fraction of bean seeds that sprouted is the dependent variable . the dependent variable ( fraction of seeds sprouting ) depends on the independent variable ( the amount of water ) , and not vice versa . experimental data ( singular : datum ) are observations made during the experiment . in this case , the data we collected were the number of bean sprouts in each pot after a week . variability and repetition out of the ten watered bean seeds , only nine came up . what happened to the tenth seed ? that seed may have been dead , unhealthy , or just slow to sprout . especially in biology ( which studies complex , living things ) , there is often variation in the material used for an experiment – here , the bean seeds – that the experimenter can not see . because of this potential for variation , biology experiments need to have a large sample size and , ideally , be repeated several times . sample size refers to the number of individual items tested in an experiment – in this case , $ 10 $ bean seeds per group . having more samples and repeating the experiment more times makes it less likely that we will reach a wrong conclusion because of random variation . biologists and other scientists also use statistical tests to help them distinguish real differences from differences due to random variation ( e.g. , when comparing experimental and control groups ) . controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ . however , a team of australian researchers hypothesized that other factors might be important too . specifically , they tested the hypothesis that high $ \text { co } _2 $ levels , which make ocean waters more acidic , might also promote bleaching $ ^2 $ . what kind of experiment would you do to test this hypothesis ? think about : what your control and experimental groups would be what your independent and dependent variables would be what results you would predict in each group have you given it a try ? non-experimental hypothesis tests some types of hypotheses ca n't be tested in controlled experiments for ethical or practical reasons . for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white . researchers suspected that high water temperature might cause bleaching , and they tested this hypothesis experimentally on a small scale ( using isolated coral fragments in tanks ) $ ^ { 3,4 } $ . what ecologists most wanted to know , however , was whether water temperature was causing bleaching for lots of different coral species in their natural setting . this broader question could not be answered experimentally , as it would n't be ethical ( or even possible ) to artificially change the water temperature surrounding entire coral reefs . instead , to test the hypothesis that natural bleaching events were caused by increases in water temperature , a team of researchers wrote a computer program to predict bleaching events based on real-time water temperature data . for example , this program would generally predict bleaching for a particular reef when the water temperature in the reef 's area exceeded its average monthly maximum by $ 1 $ $ ^\circ \text c $ or more $ ^ { 1 } $ . the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white .
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if these are all true why is n't there the same amount of coral bleaching in south east asia or the middle east compared to the antarctica ?
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introduction biologists and other scientists use the scientific method to ask questions about the natural world . the scientific method begins with an observation , which leads the scientist to ask a question . she or he then comes up with a hypothesis , a testable explanation that addresses the question . a hypothesis is n't necessarily right . instead , it 's a `` best guess , '' and the scientist must test it to see if it 's actually correct . scientists test hypotheses by making predictions : if hypothesis $ \text x $ is right , then $ \text y $ should be true . then , they do experiments or make observations to see if the predictions are correct . if they are , the hypothesis is supported . if they are n't , it may be time for a new hypothesis . how are hypotheses tested ? when possible , scientists test their hypotheses using controlled experiments . a controlled experiment is a scientific test done under controlled conditions , meaning that just one ( or a few ) factors are changed at a time , while all others are kept constant . we 'll look closely at controlled experiments in the next section . in some cases , there is no good way to test a hypothesis using a controlled experiment ( for practical or ethical reasons ) . in that case , a scientist may test a hypothesis by making predictions about patterns that should be seen in nature if the hypothesis is correct . then , she or he can collect data to see if the pattern is actually there . controlled experiments what are the key ingredients of a controlled experiment ? to illustrate , let 's consider a simple ( even silly ) example . suppose i decide to grow bean sprouts in my kitchen , near the window . i put bean seeds in a pot with soil , set them on the windowsill , and wait for them to sprout . however , after several weeks , i have no sprouts . why not ? well ... it turns out i forgot to water the seeds . so , i hypothesize that they did n't sprout due to lack of water . to test my hypothesis , i do a controlled experiment . in this experiment , i set up two identical pots . both contain ten bean seeds planted in the same type of soil , and both are placed in the same window . in fact , there is only one thing that i do differently to the two pots : one pot of seeds gets watered every afternoon . the other pot of seeds does n't get any water at all . after a week , nine out of ten seeds in the watered pot have sprouted , while none of the seeds in the dry pot have sprouted . it looks like the `` seeds need water '' hypothesis is probably correct ! let 's see how this simple example illustrates the parts of a controlled experiment . control and experimental groups there are two groups in the experiment , and they are identical except that one receives a treatment ( water ) while the other does not . the group that receives the treatment in an experiment ( here , the watered pot ) is called the experimental group , while the group that does not receive the treatment ( here , the dry pot ) is called the control group . the control group provides a baseline that lets us see if the treatment has an effect . independent and dependent variables the factor that is different between the control and experimental groups ( in this case , the amount of water ) is known as the independent variable . this variable is independent because it does not depend on what happens in the experiment . instead , it is something that the experimenter applies or chooses him/herself . in contrast , the dependent variable in an experiment is the response that 's measured to see if the treatment had an effect . in this case , the fraction of bean seeds that sprouted is the dependent variable . the dependent variable ( fraction of seeds sprouting ) depends on the independent variable ( the amount of water ) , and not vice versa . experimental data ( singular : datum ) are observations made during the experiment . in this case , the data we collected were the number of bean sprouts in each pot after a week . variability and repetition out of the ten watered bean seeds , only nine came up . what happened to the tenth seed ? that seed may have been dead , unhealthy , or just slow to sprout . especially in biology ( which studies complex , living things ) , there is often variation in the material used for an experiment – here , the bean seeds – that the experimenter can not see . because of this potential for variation , biology experiments need to have a large sample size and , ideally , be repeated several times . sample size refers to the number of individual items tested in an experiment – in this case , $ 10 $ bean seeds per group . having more samples and repeating the experiment more times makes it less likely that we will reach a wrong conclusion because of random variation . biologists and other scientists also use statistical tests to help them distinguish real differences from differences due to random variation ( e.g. , when comparing experimental and control groups ) . controlled experiment case study : $ \text { co } _2 $ and coral bleaching as a more realistic example of a controlled experiment , let 's examine a recent study on coral bleaching . corals normally have tiny photosynthetic organisms living inside of them , and bleaching happens when they leave the coral , typically due to environmental stress . the photo below shows a bleached coral in front and a healthy coral in back . a lot of research on the cause of bleaching has focused on water temperature $ ^1 $ . however , a team of australian researchers hypothesized that other factors might be important too . specifically , they tested the hypothesis that high $ \text { co } _2 $ levels , which make ocean waters more acidic , might also promote bleaching $ ^2 $ . what kind of experiment would you do to test this hypothesis ? think about : what your control and experimental groups would be what your independent and dependent variables would be what results you would predict in each group have you given it a try ? non-experimental hypothesis tests some types of hypotheses ca n't be tested in controlled experiments for ethical or practical reasons . for example , a hypothesis about viral infection ca n't be tested by dividing healthy people into two groups and infecting one group : infecting healthy people would not be safe or ethical . similarly , an ecologist studying the effects of rainfall ca n't make it rain in one part of a continent , while keeping another part dry as a control . in situations like these , biologists may use non-experimental forms of hypothesis testing . in a non-experimental hypothesis test , a researcher predicts observations or patterns that should be seen in nature if the hypothesis is correct . she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white . researchers suspected that high water temperature might cause bleaching , and they tested this hypothesis experimentally on a small scale ( using isolated coral fragments in tanks ) $ ^ { 3,4 } $ . what ecologists most wanted to know , however , was whether water temperature was causing bleaching for lots of different coral species in their natural setting . this broader question could not be answered experimentally , as it would n't be ethical ( or even possible ) to artificially change the water temperature surrounding entire coral reefs . instead , to test the hypothesis that natural bleaching events were caused by increases in water temperature , a team of researchers wrote a computer program to predict bleaching events based on real-time water temperature data . for example , this program would generally predict bleaching for a particular reef when the water temperature in the reef 's area exceeded its average monthly maximum by $ 1 $ $ ^\circ \text c $ or more $ ^ { 1 } $ . the computer program was able to predict many bleaching events weeks or even months before they were reported , including a large bleaching event in the great barrier reef in 1998 $ ^ { 1 } $ . the fact that a temperature-based model could predict bleaching events supported the hypothesis that high water temperature causes bleaching in naturally occurring coral reefs .
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she or he then collects and analyzes data , seeing whether the patterns are actually present . case study : coral bleaching and temperature a good example of hypothesis testing based on observation comes from early studies of coral bleaching . as mentioned above , bleaching is when corals lose the photosynthetic microorganisms that live inside of them , which makes them turn white .
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so both acidity and the temperature in ocean affects the coral bleaching ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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\quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations .
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can all of these answers be abstracted by doing all the algebraic manipulations beforehand and just plugging in the variables into one gigantic formula ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime .
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is there such things as three-dimensional motion ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime .
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if so is it calculated by breaking up the 3 velocities into 2d and breaking into 1d from there ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation .
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for two-dimensional projectile motion , is there ever an instance where the initial horizontal velocity is not constant ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile .
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why is the acceleration of projectiles always -g and not g ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile .
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why is g negative for projectiles ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 .
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in example 2 , why did we use tan^-1 instead of sin^-1 or cos^-1 ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile .
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at the top of the projectiles motion , does the y component of the velocity balance out with the downward pull of gravity ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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\quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations .
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if you put example 1 and 2 inside a coordinate-system - what would the x- and y-axis represent ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ .
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how is t = v_0*sin ( theta ) /g derived ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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\quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations .
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in example 2 : how is the angle on a tangent and how does it lie in a quadrant ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation .
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how does the mass , weight , and speed of a 2-d projectile affect the force of air resistance acting on the object ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ .
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wait so for the velocity , should n't it be negative since the pumpkin is displacing in the downwards direction ( assuming that standard convention is used ) ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ .
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should n't the total final velocity also be a negative , as the pumpkin is heading downward ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components .
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when throwing a ball into the air , would the initial velocity ( for both the horizontal and vertical components ) be 0 m/s ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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\quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations .
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why is acceleration left whole in some but divide by 2 in others ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) .
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how would you use the velocity to find the x and y position on the graph at a specific given time ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ .
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in the pumpkin problem why is n't the final velocity ( v ) negative ( as the pumpkin is traveling downwards ) ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ?
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why was 18m used for the distance of the vertical component for the second question on the pumpkin ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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$ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground .
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in the second problem , to calculate the final velocity does he have to use the height the pumpkin reaches before starting falling down as displacement ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward .
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technically , would n't the bullet that was fired take longer to reach the ground since it travels partway around earth which is a sphere ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components .
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is it possible to calculate initial velocity and time in air , knowing the vertical and horizontal displacement , along with the angle of launch ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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$ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ?
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what if the cannon launches the pumpkin with an angle of 230* ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime .
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but is there no influence from the centrifugal power ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity .
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so , a projectile may or may not have an initial vertical component ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ .
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when do i consider a distance to be positive or negative ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity .
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how do you derive the range formula , time in air formula , and maximum height formula ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air .
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and how would one apply them when an projectile was being launched from a given height that was different from the height that the projectile will land ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity .
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if the initial angle is below the horizon , does that mean the initial velocity would be negative ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) .
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with what speed must the animal leave the ground to reach that height ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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$ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground .
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`` this speed of v=21 m/s is the magnitude of the final velocity '' so what would be the actual velocity ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air .
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why are we not taking into account the positive distance the projectile traveled into the total displacement value ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions .
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this value is 4.12m above the cliff , should n't we add this value to the 18 meters given to give a total displacement value of -22.12m ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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\quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations .
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my question is even without the complicated variables does all of the rules of the formula still apply even to the most basic definitions ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ .
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why we do n't consider the maximum height in last example ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ?
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so , when the pumpkin is thrown up it reaches it 's max.height , where vy is zero , only after it travels downward with initial velocity zero - > it means it travels further distance and the velocity before it strikes the ground should be higher because pumpkin has more time to accelerate , no ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile .
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in the second question what will be the distance travelled by the pumpkin in x direction ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward .
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if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time , but although negligible , since the fired bullet traveled a greater distance , would an observer record an increase in the bullet 's speed ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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$ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) .
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hello , i have a question about the height , when you threw this pumpkin into the sky with an angle , the height is higher than 18m , because before the pumpkin starts dropping to the ground , it has a short period of time inclining to the sky and then the vertical velocity will reach 0m/s and then it will start dropping to the ground , do n't we need to calculate this ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables .
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how far will the rock go before it lands on the ground ?
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what 's a 2d projectile ? in a fructose induced rage , you decide to throw a lime at an angle through the air . it takes a path through space as shown by the curved , dashed line in the diagram below . the lime in this case is considered to be a two-dimensional projectile since it 's flying both vertically and horizontally through the air , and it 's only under the influence of gravity . since the gravitational force pulls downward , gravity will only affect the vertical component of the velocity $ \red { v_y } $ of the lime . the horizontal component of the velocity $ \blue { v_x } $ will remain unaffected and stay constant as the lime moves along its path . try sliding the dot in the diagram below to see that the vertical velocity $ \red { v_y } $ changes , but the horizontal velocity $ \blue { v_x } $ remains constant . concept check : at the maximum height of the lime 's trajectory , what is the value of the vertical component of velocity ? how do we handle 2d projectile motion mathematically ? one of the easiest ways to deal with 2d projectile motion is to just analyze the motion in each direction separately . in other words , we will use one set of equations to describe the horizontal motion of the lime , and another set of equations to describe the vertical motion of the lime . this turns a single difficult 2d problem into two simpler 1d problems . we 're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime . similarly , throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime . in other words , if you fire a bullet horizontally and drop a bullet at the same time , they will hit the ground at the same time . horizontal direction : there 's no acceleration in the horizontal direction since gravity does not pull projectiles sideways , only downward . air resistance would cause a horizontal acceleration , slowing the horizontal motion , but since we 're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile . so for the horizontal direction we can use the following equation , $ \large \delta x=v_xt \quad $ note : be sure to only plug horizontal variables into this horizontal equation . if we know two of the variables in this equation we can solve for the remaining unknown variable . vertical direction : two-dimensional projectiles experience a constant downward acceleration due to gravity $ a_y=-9.8 \dfrac { \text { m } } { \text { s } ^2 } $ . since the vertical acceleration is constant , we can solve for a vertical variable with one of the four kinematic formulas which are shown below . $ \large 1 . \quad v_y=v_ { 0y } +a_yt $ $ \large 2 . \quad { \delta y } = ( \dfrac { v_y+v_ { 0y } } { 2 } ) t $ $ \large 3 . \quad \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 $ $ \large 4 . \quad v_y^2=v_ { 0y } ^2+2a_y\delta y $ be sure to only plug vertical variables into these vertical equations . if we know three of the variables in these equations we can solve for any of the remaining unknown variables . note : for a given process , the time interval $ t $ has the same value for the vertical and horizontal equations . this means that if we ever solve for the time $ t $ , we can plug that time $ t $ into the equations for either the vertical or horizontal directions . this strategy is used in many problems . often , one has to solve for time $ t $ using the vertical equations , then plug that time into the horizontal equation ( or vice versa ) . what 's confusing about 2d projectile motion ? many times , people try to substitute vertical components into a horizontal equation , or vice versa . analyzing each direction ( horizontal and vertical ) of a projectile independently only works if you keep the different directions ( $ x $ or $ y $ ) in their own separate equations . initial velocities that are directed diagonally will have to be broken into vertical and horizontal components . people sometimes have a hard time breaking a velocity vector into vertical and horizontal components . see this article for help with the trigonometry you use to break vectors into components . when a projectile is shot horizontally , the initial vertical velocity is zero $ \red { v_ { 0y } =0 } $ ( see example 1 below ) . many learners have a hard time understanding that an object can start with a horizontal component of velocity , yet have zero vertical component of velocity . what do solved examples involving 2d projectile motion look like ? example 1 : horizontally launched water balloon a water balloon is thrown horizontally with a speed of $ v_0=8.31 \dfrac { \text m } { \text { s } } $ from the roof of a building of height $ h=23.0\text { m } $ . how far does the balloon travel horizontally before striking the ground ? we can start by drawing a diagram that includes the given variables . once we find the time of flight $ t $ , we 'll be able to solve for the horizontal displacement using $ \delta x=v_xt $ . to solve for time , consider the fact that we know three variables in the vertical direction ( $ \delta y=-23.0\text { m } $ , $ v_ { 0y } =0 $ , $ a=-9.8\dfrac { \text m } { \text { s } ^2 } $ ) . so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ . $ \delta x=v_x t \quad \text { ( use the equation for the horizontal displacement ) } $ $ \delta x= ( 8.31 \dfrac { \text m } { \text s } ) ( 2.17\text { s } ) \quad \text { ( plug in the time of flight and } v_x ) $ $ \delta x=18.0 \text { m } \quad \text { ( calculate and celebrate ) } $ so the water balloon struck the ground $ 18.0 \text { m } $ horizontally from the edge of the building . example 2 : pumpkin launched at an angle an air cannon is used to launch a pumpkin off a cliff of height $ h=18.0 \text { m } $ with an initial speed $ v_0=11.4 \dfrac { \text m } { \text s } $ at an angle of $ \theta=52.1^\circ $ as seen in the diagram below . what is the speed of the pumpkin right before it hits the ground ? we 'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ( $ v_x $ and $ v_y $ ) . before we can do that , we 'll have to find the components of the initial velocity ( $ v_ { 0x } $ and $ v_ { 0y } $ ) using the definitions of sine and cosine . $ \text { cos } \theta=\dfrac { \text { adjacent } } { \text { hypotenuse } } =\dfrac { v_ { 0x } } { v_0 } \quad \text { ( use the definition of cosine ) } $ $ v_ { 0x } =v_0 \text { cos } \theta \quad \text { ( solve for } v_ { 0x } ) $ $ v_ { 0x } = ( 11.4 \dfrac { \text m } { \text s } ) \text { cos } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0x } ) $ ( note : if that seemed like indecipherable mathematical witchcraft , check out this article for help with breaking vectors into components . ) this value we found for the horizontal component of the initial velocity $ v_ { 0x } =7.00 \dfrac { \text m } { \text s } $ will also be the horizontal component of the final velocity $ v_x=7.00 \dfrac { \text m } { \text s } $ since the horizontal component of the velocity remains constant for the entire flight ( assuming there is no air resistance ) . to find the vertical component of the initial velocity we 'll use the same procedure as above but with sine instead of cosine . $ \text { sin } \theta=\dfrac { \text { opposite } } { \text { hypotenuse } } =\dfrac { v_ { 0y } } { v_0 } \quad \text { ( use the definition of sine ) } $ $ v_ { 0y } =v_0 \text { sin } \theta \quad \text { ( solve for } v_ { 0y } ) $ $ v_ { 0y } = ( 11.4 \dfrac { \text m } { \text s } ) \text { sin } ( 52.1^\circ ) \quad \text { ( plug in numerical values ) } $ $ v_ { 0y } =9.00 \dfrac { \text m } { \text s } \quad \text { ( calculate to find } v_ { 0y } ) $ since the vertical component of velocity $ v_y $ changes for a projectile as it moves through the air we 'll have to solve for the vertical component of the final velocity $ v_y $ using a kinematic formula for the vertical direction . since we do n't know the time of flight $ t $ , and we were n't asked to find the time $ t $ , we 'll use the vertical kinematic formula that does n't include time $ t $ . $ v_y^2=v_ { 0y } ^2+2a_y\delta y \quad \text { ( use the kinematic formula that does n't include time ) } $ $ v_y^2= ( 9.00 \dfrac { \text m } { \text s } ) ^2+2 ( -9.8\dfrac { \text m } { \text { s } ^2 } ) ( -18 \text { m } ) \quad \text { ( plug in known values ) } $ $ v_y^2=434\dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v_y=\pm\sqrt { 434\dfrac { \text { m } ^2 } { \text { s } ^2 } } =\pm20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ $ v_y=-20.8 \dfrac { \text { m } } { \text { s } } \quad \text { ( choose the negative root since the pumpkin will be heading downwards ) } $ now that we know the horizontal and vertical components of the final velocity , we can use the pythagorean theorem to solve for the final speed ( i.e . magnitude of the final velocity ) . $ v^2=v_x^2+v_y^2 \quad \text { ( use the pythagorean theorem ) } $ $ v^2= ( 7.00 \dfrac { \text m } { \text s } ) ^2+ ( -20.8 \dfrac { \text { m } } { \text { s } } ) ^2 \quad \text { ( plug in horizontal and vertical components of the final velocity ) } $ $ v^2=482 \dfrac { \text { m } ^2 } { \text { s } ^2 } \quad \text { ( calculate ) } $ $ v=21.9 \dfrac { \text { m } } { \text { s } } \quad \text { ( take a square root ) } $ this speed of $ v=21.9 \dfrac { \text { m } } { \text { s } } $ is the magnitude of the final velocity of the pumpkin right before it hits the ground . the relationship between the final velocity and its components is shown in the diagram below . we could also solve for the angle $ \phi $ of the final velocity using the definition of tangent . $ \text { tan } \phi=\dfrac { \text { opposite } } { \text { adjacent } } =\dfrac { v_y } { v_x } $ $ \text { tan } \phi=\dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } $ now taking inverse tangent of both sides we get , $ tan^ { -1 } ( \text { tan } \phi ) =tan^ { -1 } ( \dfrac { 20.8\dfrac { \text m } { \text s } } { 7.00\dfrac { \text m } { \text s } } ) $ the left hand side just becomes $ \phi $ , and we can find the value of the right hand side by plugging into a calculator to get , $ \phi=71.4^\circ $
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so we 'll use a kinematic formula in the vertical direction to solve for time $ t $ . we do n't know the final velocity $ v_y $ , and we are n't asked for the final velocity $ v_y $ so we 'll use the vertical kinematic formula that does n't include final velocity . $ \delta y=v_ { 0y } t+\dfrac { 1 } { 2 } a_yt^2 \quad \text { ( use the vertical kinematic formula that does n't include final velocity ) } $ $ -h= ( 0 ) t+\dfrac { 1 } { 2 } ( -g ) t^2 \quad \text { ( plug in known vertical values } ) $ $ t= \sqrt { \dfrac { 2h } { g } } \quad \text { ( solve symbolically for time t } ) $ $ t= \sqrt { \dfrac { 2 ( -23.0\text { m } ) } { -9.8\dfrac { \text m } { \text { s } ^2 } } } =2.17 \text { s } \quad \text { ( plug in numerical values and find the time of flight } ) $ now we need to plug this time $ t $ into the equation for the horizontal direction to find horizontal displacement $ \delta x $ .
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in the last solved problem , the final velocity is positive , should n't it be negative since it 's direction is downwards ?
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background multivariable functions parametric functions , one parameter what we 're building to you can visualize a function with a two-dimensional input and a three-dimensional output by plotting all the output points corresponding to some region of the input space . this results in a surface , known as a parametric surface . the process of going the other way around , starting with a surface in space and trying to find a function that `` draws '' this surface , is known as parameterizing the surface . in general , this is a tricky thing to do . quick review of one-parameter functions in the last article , i talked about visualizing functions with a one-dimensional input and a two-dimensional output . for example : $ \displaystyle f ( t ) = \left [ \begin { array } { c } t\cos ( t ) \ \sin ( t ) \end { array } \right ] $ i talked about how because the output space has more dimensions than the input space , you can get a good feel for the function just by seeing which points in the output space are `` hit '' by the function as the input $ t $ ranges over some set of values . when a function is interpreted this way , it is known as a parametric function , and its input $ t $ is called the parameter two parameters we can do something very similar with functions that have a two-dimensional input and a three-dimensional output . $ \displaystyle f ( s , t ) = \left [ \begin { array } { c } t^3 - st \ s-t \ s+t \end { array } \right ] $ both input coordinates $ s $ and $ t $ will be known as the parameters , and you are about to see how this function draws a surface in three-dimensional space . the first step to representing a function like this is to specify a range for the input , such as $ \begin { align } \quad 0 & lt ; & amp ; s & lt ; 3 \ -2 & lt ; & amp ; t & lt ; 2 \end { align } $ here 's what that region looks like in the input space . next , we consider all possible outputs of the function in that range . input $ ( s , t ) $ | output $ ( t^3 -st , s-t , s+t ) $ -|- $ ( 0 , 0 ) $ | $ ( 0 , 0 , 0 ) $ $ ( 1 , 0 ) $ | $ ( 1 , 1 , 1 ) $ $ ( 2 , 1 ) $ | $ ( 6 , 1 , 3 ) $ $ \quad\vdots $ | $ \quad\ ; \vdots $ okay , so we do n't literally write out all possible outputs , since , you know , that involves infinitely many things . in principle , though , our goal is to represent all those values . since the function spits out a three-coordinate output , we visualize this output in three-dimensional space . the following animation shows what it looks like as the points $ ( s , t ) $ in the parameter space move to the corresponding output $ f ( s , t ) $ in three-dimensional space : the resulting surface in three-dimensional space is called a parametric surface . warning : surfaces like these can be confused with the graphs of functions that have a two-dimensional input and a one-dimensional output since those are also drawn as surfaces in three-dimensional space . but these parametric functions have a very different flavor . they have a two-dimensional input and a three-dimensional output . notice , this means graphing them would require five dimensions ! parameterizing a surface one of the best ways to get a feel for parametric functions is to start with a surface that you want to describe , then try to find a function that will draw this as a parametric surface . this is also a necessary skill when you start learning about surface integrals later on in multivariable calculus . be warned , though , parameterizing surfaces is not easy . in the following example , we will parameterize a torus , the fancy word for the surface of a doughnut . as surfaces go , a torus is a relatively simple example , but it still takes some serious effort . example : parameterize a torus ( doughnut ) consider the surface pictured above . you can think of it as a doughnut shape , or perhaps just as the glaze on the doughnut , since we do n't care about the filling . our goal right now is to find a function with a two-dimensional input , and a three-dimensional output , such that this doughnut shape is the output . we imagine `` drawing '' the surface , although one does not simply draw a surface with a pencil and paper the way we can draw a curve . instead , our strategy will be to draw each circular slice of the torus . to see what i mean , here is a sample of those circular slices ( drawn in blue ) : i also drew a large red circle on the $ xy $ -plane running through the centers of each of these slices . this is not part of the torus , but will be a useful reference point for the end goal of drawing each blue slice . in a real problem , the radius of the red circle might be given to you , as would the radius of each blue circular slice . for now , let 's choose arbitrarily that the radius of the red circle is $ 3 $ , and the radius of each blue slice is $ 1 $ , with the understanding that choosing different values would give different toruses ( torii ? torotes ? ) . core idea : we will describe each point on the torus as the sum of two vectors : a vector $ \vec { \textbf { c } } $ from the origin to a point on the red circle . to specify which point on the red circle , we will make this a vector-valued function that depends on a parameter $ t $ . as the value of $ t $ changes , the point on the red circle described by $ \vec { \textbf { c } } ( t ) $ will change . a vector $ \vec { \textbf { d } } $ from that point on the red circle to a point on the corresponding `` slice '' of the torus . the direction this vector points will depend on which point of the red circle it is anchored to , so the value of $ \vec { \textbf { d } } $ should depend on the parameter $ t $ used to describe points of the red circle . furthermore , we will use a second parameter $ u $ to determine which part of the blue torus slice $ \vec { \textbf { d } } $ points to . this means points on the torus will each be described as a sum . $ \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) $ ( if you are unfamiliar with the tip-to-tail method of adding vectors , consider reviewing this video ) . why this strategy ? the idea here is that we do n't immediately know how to define points on a torus , but we do know how to define circles . since the big red circle is flat on the $ xy $ -plane , and has radius $ 3 $ , we can parameterize it as follows : $ \begin { align } \quad \vec { \textbf { c } } ( t ) = 3 \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] = 3 \cos ( t ) \hat { \textbf { i } } + 3 \sin ( t ) \hat { \textbf { j } } + 0 \hat { \textbf { k } } \end { align } $ now , the vector-valued function $ \vec { \textbf { d } } ( u , t ) $ should also describe a circle , but it 's a bit trickier . the ( blue ) circular slice of the torus we want $ \vec { \textbf { d } } ( u , t ) $ to draw is at an angle . how do you draw a circle which is sitting at an angle in three-dimensional space ? well , let 's start from what we know . we know that in two dimensions , a unit circle centered at the origin can be described with the parametric function $ \begin { align } \quad g ( u ) = \left [ \begin { array } { a } \cos ( u ) \ \sin ( u ) \end { array } \right ] = \cos ( u ) \hat { \textbf { i } } + \sin ( u ) \hat { \textbf { j } } \end { align } $ for our desired blue circular slice , we do something similar , but we exchange $ \hat { \textbf { i } } $ and $ \hat { \textbf { j } } $ for different unit vectors . take a look at this picture : instead of the `` sideways '' direction being $ \hat { \textbf { i } } $ , the unit vector in the $ x $ -direction , we think of it as the unit vector pointing away from the origin , which we 'll call $ \hat { \textbf { v } } $ . actually , since that direction might depend on where we start , $ \hat { \textbf { v } } $ should be a vector-valued function dependent on the parameter $ t $ , so we write it as $ \hat { \textbf { v } } ( t ) $ . similarly the `` upward '' direction is no longer $ \hat { \textbf { j } } $ , but $ \hat { \textbf { k } } $ , the unit vector in the $ z $ -direction . therefore , the parameterization of the circular slice should look something like this : $ \begin { align } \quad \vec { \textbf { d } } ( u , t ) = \cos ( u ) \hat { \textbf { v } } ( t ) + \sin ( u ) \hat { \textbf { k } } \end { align } $ this of course leaves us with a question : what is the formula for $ \hat { \textbf { v } } ( t ) $ ? looking at the picture , the direction away from the origin is also described by $ \vec { \textbf { c } } ( t ) $ , so the formula for $ \hat { \textbf { v } } ( t ) $ should be the same as that for $ \vec { \textbf { c } } ( t ) $ , but scaled down to be a unit vector . $ \begin { align } \quad \vec { \textbf { c } } ( t ) = 3 & amp ; \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] \quad \leftarrow \small { \gray { \text { not a unit vector } } } \ \downarrow & amp ; \ \hat { \textbf { v } } ( t ) = & amp ; \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] \quad \leftarrow \small { \gray { \text { unit vector } } } \end { align } $ this means our full expression for $ \vec { \textbf { d } } ( u , t ) $ is $ \begin { align } \quad \vec { \textbf { d } } ( u , t ) & amp ; = \cos ( u ) \hat { \textbf { v } } ( t ) + \sin ( u ) \hat { \textbf { k } } \ & amp ; = \cos ( u ) \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] + \sin ( u ) \left [ \begin { array } { c } 0 \ 0 \ 1 \end { array } \right ] = \left [ \begin { array } { c } \cos ( u ) \cos ( t ) \ \cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] \end { align } $ bring it on home remember , the whole reason we defined $ \vec { \textbf { d } } ( u , t ) $ and $ \vec { \textbf { c } } ( t ) $ was to describe each point on the torus as $ \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) $ . putting this together , we have the following vector-valued two-parameter function : $ \begin { align } \quad \vec { f } ( u , t ) & amp ; = \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) \ & amp ; = 3\left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] + \left [ \begin { array } { c } \cos ( u ) \cos ( t ) \ \cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] \ & amp ; = \bluee { \left [ \begin { array } { c } 3\cos ( t ) + \cos ( u ) \cos ( t ) \ 3\sin ( t ) +\cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] } \end { align } $ as $ u $ ranges from $ 0 $ to $ 2\pi $ , the output of this function $ \vec { f } ( u , t ) $ will trace one of the blue slices , and as $ t $ ranges from $ 0 $ to $ 2\pi $ , the slices themselves will trace out the entire torus . here 's what it might look like if we take the points from the parameter space where $ 0 \le u \le 2\pi $ and $ 0 \le t \le 2\pi $ , and watch them move to the output of our function $ \vec { f } ( u , t ) $ : summary you can visualize a function with a two-dimensional input and a three-dimensional output by plotting all the output points corresponding to some region of the input space . this results in a surface , known as a parametric surface . the process of going the other way around , starting with a surface in space and trying to find a function that `` draws '' this surface , is known as parameterizing the surface . in general , this is a tricky thing to do .
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$ \begin { align } \quad \vec { \textbf { c } } ( t ) = 3 & amp ; \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] \quad \leftarrow \small { \gray { \text { not a unit vector } } } \ \downarrow & amp ; \ \hat { \textbf { v } } ( t ) = & amp ; \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] \quad \leftarrow \small { \gray { \text { unit vector } } } \end { align } $ this means our full expression for $ \vec { \textbf { d } } ( u , t ) $ is $ \begin { align } \quad \vec { \textbf { d } } ( u , t ) & amp ; = \cos ( u ) \hat { \textbf { v } } ( t ) + \sin ( u ) \hat { \textbf { k } } \ & amp ; = \cos ( u ) \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] + \sin ( u ) \left [ \begin { array } { c } 0 \ 0 \ 1 \end { array } \right ] = \left [ \begin { array } { c } \cos ( u ) \cos ( t ) \ \cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] \end { align } $ bring it on home remember , the whole reason we defined $ \vec { \textbf { d } } ( u , t ) $ and $ \vec { \textbf { c } } ( t ) $ was to describe each point on the torus as $ \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) $ . putting this together , we have the following vector-valued two-parameter function : $ \begin { align } \quad \vec { f } ( u , t ) & amp ; = \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) \ & amp ; = 3\left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] + \left [ \begin { array } { c } \cos ( u ) \cos ( t ) \ \cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] \ & amp ; = \bluee { \left [ \begin { array } { c } 3\cos ( t ) + \cos ( u ) \cos ( t ) \ 3\sin ( t ) +\cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] } \end { align } $ as $ u $ ranges from $ 0 $ to $ 2\pi $ , the output of this function $ \vec { f } ( u , t ) $ will trace one of the blue slices , and as $ t $ ranges from $ 0 $ to $ 2\pi $ , the slices themselves will trace out the entire torus . here 's what it might look like if we take the points from the parameter space where $ 0 \le u \le 2\pi $ and $ 0 \le t \le 2\pi $ , and watch them move to the output of our function $ \vec { f } ( u , t ) $ : summary you can visualize a function with a two-dimensional input and a three-dimensional output by plotting all the output points corresponding to some region of the input space .
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why is cos ( u ) multiplied by sin ( t ) ?
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background multivariable functions parametric functions , one parameter what we 're building to you can visualize a function with a two-dimensional input and a three-dimensional output by plotting all the output points corresponding to some region of the input space . this results in a surface , known as a parametric surface . the process of going the other way around , starting with a surface in space and trying to find a function that `` draws '' this surface , is known as parameterizing the surface . in general , this is a tricky thing to do . quick review of one-parameter functions in the last article , i talked about visualizing functions with a one-dimensional input and a two-dimensional output . for example : $ \displaystyle f ( t ) = \left [ \begin { array } { c } t\cos ( t ) \ \sin ( t ) \end { array } \right ] $ i talked about how because the output space has more dimensions than the input space , you can get a good feel for the function just by seeing which points in the output space are `` hit '' by the function as the input $ t $ ranges over some set of values . when a function is interpreted this way , it is known as a parametric function , and its input $ t $ is called the parameter two parameters we can do something very similar with functions that have a two-dimensional input and a three-dimensional output . $ \displaystyle f ( s , t ) = \left [ \begin { array } { c } t^3 - st \ s-t \ s+t \end { array } \right ] $ both input coordinates $ s $ and $ t $ will be known as the parameters , and you are about to see how this function draws a surface in three-dimensional space . the first step to representing a function like this is to specify a range for the input , such as $ \begin { align } \quad 0 & lt ; & amp ; s & lt ; 3 \ -2 & lt ; & amp ; t & lt ; 2 \end { align } $ here 's what that region looks like in the input space . next , we consider all possible outputs of the function in that range . input $ ( s , t ) $ | output $ ( t^3 -st , s-t , s+t ) $ -|- $ ( 0 , 0 ) $ | $ ( 0 , 0 , 0 ) $ $ ( 1 , 0 ) $ | $ ( 1 , 1 , 1 ) $ $ ( 2 , 1 ) $ | $ ( 6 , 1 , 3 ) $ $ \quad\vdots $ | $ \quad\ ; \vdots $ okay , so we do n't literally write out all possible outputs , since , you know , that involves infinitely many things . in principle , though , our goal is to represent all those values . since the function spits out a three-coordinate output , we visualize this output in three-dimensional space . the following animation shows what it looks like as the points $ ( s , t ) $ in the parameter space move to the corresponding output $ f ( s , t ) $ in three-dimensional space : the resulting surface in three-dimensional space is called a parametric surface . warning : surfaces like these can be confused with the graphs of functions that have a two-dimensional input and a one-dimensional output since those are also drawn as surfaces in three-dimensional space . but these parametric functions have a very different flavor . they have a two-dimensional input and a three-dimensional output . notice , this means graphing them would require five dimensions ! parameterizing a surface one of the best ways to get a feel for parametric functions is to start with a surface that you want to describe , then try to find a function that will draw this as a parametric surface . this is also a necessary skill when you start learning about surface integrals later on in multivariable calculus . be warned , though , parameterizing surfaces is not easy . in the following example , we will parameterize a torus , the fancy word for the surface of a doughnut . as surfaces go , a torus is a relatively simple example , but it still takes some serious effort . example : parameterize a torus ( doughnut ) consider the surface pictured above . you can think of it as a doughnut shape , or perhaps just as the glaze on the doughnut , since we do n't care about the filling . our goal right now is to find a function with a two-dimensional input , and a three-dimensional output , such that this doughnut shape is the output . we imagine `` drawing '' the surface , although one does not simply draw a surface with a pencil and paper the way we can draw a curve . instead , our strategy will be to draw each circular slice of the torus . to see what i mean , here is a sample of those circular slices ( drawn in blue ) : i also drew a large red circle on the $ xy $ -plane running through the centers of each of these slices . this is not part of the torus , but will be a useful reference point for the end goal of drawing each blue slice . in a real problem , the radius of the red circle might be given to you , as would the radius of each blue circular slice . for now , let 's choose arbitrarily that the radius of the red circle is $ 3 $ , and the radius of each blue slice is $ 1 $ , with the understanding that choosing different values would give different toruses ( torii ? torotes ? ) . core idea : we will describe each point on the torus as the sum of two vectors : a vector $ \vec { \textbf { c } } $ from the origin to a point on the red circle . to specify which point on the red circle , we will make this a vector-valued function that depends on a parameter $ t $ . as the value of $ t $ changes , the point on the red circle described by $ \vec { \textbf { c } } ( t ) $ will change . a vector $ \vec { \textbf { d } } $ from that point on the red circle to a point on the corresponding `` slice '' of the torus . the direction this vector points will depend on which point of the red circle it is anchored to , so the value of $ \vec { \textbf { d } } $ should depend on the parameter $ t $ used to describe points of the red circle . furthermore , we will use a second parameter $ u $ to determine which part of the blue torus slice $ \vec { \textbf { d } } $ points to . this means points on the torus will each be described as a sum . $ \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) $ ( if you are unfamiliar with the tip-to-tail method of adding vectors , consider reviewing this video ) . why this strategy ? the idea here is that we do n't immediately know how to define points on a torus , but we do know how to define circles . since the big red circle is flat on the $ xy $ -plane , and has radius $ 3 $ , we can parameterize it as follows : $ \begin { align } \quad \vec { \textbf { c } } ( t ) = 3 \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] = 3 \cos ( t ) \hat { \textbf { i } } + 3 \sin ( t ) \hat { \textbf { j } } + 0 \hat { \textbf { k } } \end { align } $ now , the vector-valued function $ \vec { \textbf { d } } ( u , t ) $ should also describe a circle , but it 's a bit trickier . the ( blue ) circular slice of the torus we want $ \vec { \textbf { d } } ( u , t ) $ to draw is at an angle . how do you draw a circle which is sitting at an angle in three-dimensional space ? well , let 's start from what we know . we know that in two dimensions , a unit circle centered at the origin can be described with the parametric function $ \begin { align } \quad g ( u ) = \left [ \begin { array } { a } \cos ( u ) \ \sin ( u ) \end { array } \right ] = \cos ( u ) \hat { \textbf { i } } + \sin ( u ) \hat { \textbf { j } } \end { align } $ for our desired blue circular slice , we do something similar , but we exchange $ \hat { \textbf { i } } $ and $ \hat { \textbf { j } } $ for different unit vectors . take a look at this picture : instead of the `` sideways '' direction being $ \hat { \textbf { i } } $ , the unit vector in the $ x $ -direction , we think of it as the unit vector pointing away from the origin , which we 'll call $ \hat { \textbf { v } } $ . actually , since that direction might depend on where we start , $ \hat { \textbf { v } } $ should be a vector-valued function dependent on the parameter $ t $ , so we write it as $ \hat { \textbf { v } } ( t ) $ . similarly the `` upward '' direction is no longer $ \hat { \textbf { j } } $ , but $ \hat { \textbf { k } } $ , the unit vector in the $ z $ -direction . therefore , the parameterization of the circular slice should look something like this : $ \begin { align } \quad \vec { \textbf { d } } ( u , t ) = \cos ( u ) \hat { \textbf { v } } ( t ) + \sin ( u ) \hat { \textbf { k } } \end { align } $ this of course leaves us with a question : what is the formula for $ \hat { \textbf { v } } ( t ) $ ? looking at the picture , the direction away from the origin is also described by $ \vec { \textbf { c } } ( t ) $ , so the formula for $ \hat { \textbf { v } } ( t ) $ should be the same as that for $ \vec { \textbf { c } } ( t ) $ , but scaled down to be a unit vector . $ \begin { align } \quad \vec { \textbf { c } } ( t ) = 3 & amp ; \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] \quad \leftarrow \small { \gray { \text { not a unit vector } } } \ \downarrow & amp ; \ \hat { \textbf { v } } ( t ) = & amp ; \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] \quad \leftarrow \small { \gray { \text { unit vector } } } \end { align } $ this means our full expression for $ \vec { \textbf { d } } ( u , t ) $ is $ \begin { align } \quad \vec { \textbf { d } } ( u , t ) & amp ; = \cos ( u ) \hat { \textbf { v } } ( t ) + \sin ( u ) \hat { \textbf { k } } \ & amp ; = \cos ( u ) \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] + \sin ( u ) \left [ \begin { array } { c } 0 \ 0 \ 1 \end { array } \right ] = \left [ \begin { array } { c } \cos ( u ) \cos ( t ) \ \cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] \end { align } $ bring it on home remember , the whole reason we defined $ \vec { \textbf { d } } ( u , t ) $ and $ \vec { \textbf { c } } ( t ) $ was to describe each point on the torus as $ \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) $ . putting this together , we have the following vector-valued two-parameter function : $ \begin { align } \quad \vec { f } ( u , t ) & amp ; = \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) \ & amp ; = 3\left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] + \left [ \begin { array } { c } \cos ( u ) \cos ( t ) \ \cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] \ & amp ; = \bluee { \left [ \begin { array } { c } 3\cos ( t ) + \cos ( u ) \cos ( t ) \ 3\sin ( t ) +\cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] } \end { align } $ as $ u $ ranges from $ 0 $ to $ 2\pi $ , the output of this function $ \vec { f } ( u , t ) $ will trace one of the blue slices , and as $ t $ ranges from $ 0 $ to $ 2\pi $ , the slices themselves will trace out the entire torus . here 's what it might look like if we take the points from the parameter space where $ 0 \le u \le 2\pi $ and $ 0 \le t \le 2\pi $ , and watch them move to the output of our function $ \vec { f } ( u , t ) $ : summary you can visualize a function with a two-dimensional input and a three-dimensional output by plotting all the output points corresponding to some region of the input space . this results in a surface , known as a parametric surface . the process of going the other way around , starting with a surface in space and trying to find a function that `` draws '' this surface , is known as parameterizing the surface . in general , this is a tricky thing to do .
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background multivariable functions parametric functions , one parameter what we 're building to you can visualize a function with a two-dimensional input and a three-dimensional output by plotting all the output points corresponding to some region of the input space . this results in a surface , known as a parametric surface .
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how were all animations created ?
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background multivariable functions parametric functions , one parameter what we 're building to you can visualize a function with a two-dimensional input and a three-dimensional output by plotting all the output points corresponding to some region of the input space . this results in a surface , known as a parametric surface . the process of going the other way around , starting with a surface in space and trying to find a function that `` draws '' this surface , is known as parameterizing the surface . in general , this is a tricky thing to do . quick review of one-parameter functions in the last article , i talked about visualizing functions with a one-dimensional input and a two-dimensional output . for example : $ \displaystyle f ( t ) = \left [ \begin { array } { c } t\cos ( t ) \ \sin ( t ) \end { array } \right ] $ i talked about how because the output space has more dimensions than the input space , you can get a good feel for the function just by seeing which points in the output space are `` hit '' by the function as the input $ t $ ranges over some set of values . when a function is interpreted this way , it is known as a parametric function , and its input $ t $ is called the parameter two parameters we can do something very similar with functions that have a two-dimensional input and a three-dimensional output . $ \displaystyle f ( s , t ) = \left [ \begin { array } { c } t^3 - st \ s-t \ s+t \end { array } \right ] $ both input coordinates $ s $ and $ t $ will be known as the parameters , and you are about to see how this function draws a surface in three-dimensional space . the first step to representing a function like this is to specify a range for the input , such as $ \begin { align } \quad 0 & lt ; & amp ; s & lt ; 3 \ -2 & lt ; & amp ; t & lt ; 2 \end { align } $ here 's what that region looks like in the input space . next , we consider all possible outputs of the function in that range . input $ ( s , t ) $ | output $ ( t^3 -st , s-t , s+t ) $ -|- $ ( 0 , 0 ) $ | $ ( 0 , 0 , 0 ) $ $ ( 1 , 0 ) $ | $ ( 1 , 1 , 1 ) $ $ ( 2 , 1 ) $ | $ ( 6 , 1 , 3 ) $ $ \quad\vdots $ | $ \quad\ ; \vdots $ okay , so we do n't literally write out all possible outputs , since , you know , that involves infinitely many things . in principle , though , our goal is to represent all those values . since the function spits out a three-coordinate output , we visualize this output in three-dimensional space . the following animation shows what it looks like as the points $ ( s , t ) $ in the parameter space move to the corresponding output $ f ( s , t ) $ in three-dimensional space : the resulting surface in three-dimensional space is called a parametric surface . warning : surfaces like these can be confused with the graphs of functions that have a two-dimensional input and a one-dimensional output since those are also drawn as surfaces in three-dimensional space . but these parametric functions have a very different flavor . they have a two-dimensional input and a three-dimensional output . notice , this means graphing them would require five dimensions ! parameterizing a surface one of the best ways to get a feel for parametric functions is to start with a surface that you want to describe , then try to find a function that will draw this as a parametric surface . this is also a necessary skill when you start learning about surface integrals later on in multivariable calculus . be warned , though , parameterizing surfaces is not easy . in the following example , we will parameterize a torus , the fancy word for the surface of a doughnut . as surfaces go , a torus is a relatively simple example , but it still takes some serious effort . example : parameterize a torus ( doughnut ) consider the surface pictured above . you can think of it as a doughnut shape , or perhaps just as the glaze on the doughnut , since we do n't care about the filling . our goal right now is to find a function with a two-dimensional input , and a three-dimensional output , such that this doughnut shape is the output . we imagine `` drawing '' the surface , although one does not simply draw a surface with a pencil and paper the way we can draw a curve . instead , our strategy will be to draw each circular slice of the torus . to see what i mean , here is a sample of those circular slices ( drawn in blue ) : i also drew a large red circle on the $ xy $ -plane running through the centers of each of these slices . this is not part of the torus , but will be a useful reference point for the end goal of drawing each blue slice . in a real problem , the radius of the red circle might be given to you , as would the radius of each blue circular slice . for now , let 's choose arbitrarily that the radius of the red circle is $ 3 $ , and the radius of each blue slice is $ 1 $ , with the understanding that choosing different values would give different toruses ( torii ? torotes ? ) . core idea : we will describe each point on the torus as the sum of two vectors : a vector $ \vec { \textbf { c } } $ from the origin to a point on the red circle . to specify which point on the red circle , we will make this a vector-valued function that depends on a parameter $ t $ . as the value of $ t $ changes , the point on the red circle described by $ \vec { \textbf { c } } ( t ) $ will change . a vector $ \vec { \textbf { d } } $ from that point on the red circle to a point on the corresponding `` slice '' of the torus . the direction this vector points will depend on which point of the red circle it is anchored to , so the value of $ \vec { \textbf { d } } $ should depend on the parameter $ t $ used to describe points of the red circle . furthermore , we will use a second parameter $ u $ to determine which part of the blue torus slice $ \vec { \textbf { d } } $ points to . this means points on the torus will each be described as a sum . $ \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) $ ( if you are unfamiliar with the tip-to-tail method of adding vectors , consider reviewing this video ) . why this strategy ? the idea here is that we do n't immediately know how to define points on a torus , but we do know how to define circles . since the big red circle is flat on the $ xy $ -plane , and has radius $ 3 $ , we can parameterize it as follows : $ \begin { align } \quad \vec { \textbf { c } } ( t ) = 3 \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] = 3 \cos ( t ) \hat { \textbf { i } } + 3 \sin ( t ) \hat { \textbf { j } } + 0 \hat { \textbf { k } } \end { align } $ now , the vector-valued function $ \vec { \textbf { d } } ( u , t ) $ should also describe a circle , but it 's a bit trickier . the ( blue ) circular slice of the torus we want $ \vec { \textbf { d } } ( u , t ) $ to draw is at an angle . how do you draw a circle which is sitting at an angle in three-dimensional space ? well , let 's start from what we know . we know that in two dimensions , a unit circle centered at the origin can be described with the parametric function $ \begin { align } \quad g ( u ) = \left [ \begin { array } { a } \cos ( u ) \ \sin ( u ) \end { array } \right ] = \cos ( u ) \hat { \textbf { i } } + \sin ( u ) \hat { \textbf { j } } \end { align } $ for our desired blue circular slice , we do something similar , but we exchange $ \hat { \textbf { i } } $ and $ \hat { \textbf { j } } $ for different unit vectors . take a look at this picture : instead of the `` sideways '' direction being $ \hat { \textbf { i } } $ , the unit vector in the $ x $ -direction , we think of it as the unit vector pointing away from the origin , which we 'll call $ \hat { \textbf { v } } $ . actually , since that direction might depend on where we start , $ \hat { \textbf { v } } $ should be a vector-valued function dependent on the parameter $ t $ , so we write it as $ \hat { \textbf { v } } ( t ) $ . similarly the `` upward '' direction is no longer $ \hat { \textbf { j } } $ , but $ \hat { \textbf { k } } $ , the unit vector in the $ z $ -direction . therefore , the parameterization of the circular slice should look something like this : $ \begin { align } \quad \vec { \textbf { d } } ( u , t ) = \cos ( u ) \hat { \textbf { v } } ( t ) + \sin ( u ) \hat { \textbf { k } } \end { align } $ this of course leaves us with a question : what is the formula for $ \hat { \textbf { v } } ( t ) $ ? looking at the picture , the direction away from the origin is also described by $ \vec { \textbf { c } } ( t ) $ , so the formula for $ \hat { \textbf { v } } ( t ) $ should be the same as that for $ \vec { \textbf { c } } ( t ) $ , but scaled down to be a unit vector . $ \begin { align } \quad \vec { \textbf { c } } ( t ) = 3 & amp ; \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] \quad \leftarrow \small { \gray { \text { not a unit vector } } } \ \downarrow & amp ; \ \hat { \textbf { v } } ( t ) = & amp ; \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] \quad \leftarrow \small { \gray { \text { unit vector } } } \end { align } $ this means our full expression for $ \vec { \textbf { d } } ( u , t ) $ is $ \begin { align } \quad \vec { \textbf { d } } ( u , t ) & amp ; = \cos ( u ) \hat { \textbf { v } } ( t ) + \sin ( u ) \hat { \textbf { k } } \ & amp ; = \cos ( u ) \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] + \sin ( u ) \left [ \begin { array } { c } 0 \ 0 \ 1 \end { array } \right ] = \left [ \begin { array } { c } \cos ( u ) \cos ( t ) \ \cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] \end { align } $ bring it on home remember , the whole reason we defined $ \vec { \textbf { d } } ( u , t ) $ and $ \vec { \textbf { c } } ( t ) $ was to describe each point on the torus as $ \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) $ . putting this together , we have the following vector-valued two-parameter function : $ \begin { align } \quad \vec { f } ( u , t ) & amp ; = \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) \ & amp ; = 3\left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] + \left [ \begin { array } { c } \cos ( u ) \cos ( t ) \ \cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] \ & amp ; = \bluee { \left [ \begin { array } { c } 3\cos ( t ) + \cos ( u ) \cos ( t ) \ 3\sin ( t ) +\cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] } \end { align } $ as $ u $ ranges from $ 0 $ to $ 2\pi $ , the output of this function $ \vec { f } ( u , t ) $ will trace one of the blue slices , and as $ t $ ranges from $ 0 $ to $ 2\pi $ , the slices themselves will trace out the entire torus . here 's what it might look like if we take the points from the parameter space where $ 0 \le u \le 2\pi $ and $ 0 \le t \le 2\pi $ , and watch them move to the output of our function $ \vec { f } ( u , t ) $ : summary you can visualize a function with a two-dimensional input and a three-dimensional output by plotting all the output points corresponding to some region of the input space . this results in a surface , known as a parametric surface . the process of going the other way around , starting with a surface in space and trying to find a function that `` draws '' this surface , is known as parameterizing the surface . in general , this is a tricky thing to do .
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parameterizing a surface one of the best ways to get a feel for parametric functions is to start with a surface that you want to describe , then try to find a function that will draw this as a parametric surface . this is also a necessary skill when you start learning about surface integrals later on in multivariable calculus . be warned , though , parameterizing surfaces is not easy .
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now or after i 've understood more of the multivariable calculus material , is there any way i can better understand how programming these animations is done ?
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background multivariable functions parametric functions , one parameter what we 're building to you can visualize a function with a two-dimensional input and a three-dimensional output by plotting all the output points corresponding to some region of the input space . this results in a surface , known as a parametric surface . the process of going the other way around , starting with a surface in space and trying to find a function that `` draws '' this surface , is known as parameterizing the surface . in general , this is a tricky thing to do . quick review of one-parameter functions in the last article , i talked about visualizing functions with a one-dimensional input and a two-dimensional output . for example : $ \displaystyle f ( t ) = \left [ \begin { array } { c } t\cos ( t ) \ \sin ( t ) \end { array } \right ] $ i talked about how because the output space has more dimensions than the input space , you can get a good feel for the function just by seeing which points in the output space are `` hit '' by the function as the input $ t $ ranges over some set of values . when a function is interpreted this way , it is known as a parametric function , and its input $ t $ is called the parameter two parameters we can do something very similar with functions that have a two-dimensional input and a three-dimensional output . $ \displaystyle f ( s , t ) = \left [ \begin { array } { c } t^3 - st \ s-t \ s+t \end { array } \right ] $ both input coordinates $ s $ and $ t $ will be known as the parameters , and you are about to see how this function draws a surface in three-dimensional space . the first step to representing a function like this is to specify a range for the input , such as $ \begin { align } \quad 0 & lt ; & amp ; s & lt ; 3 \ -2 & lt ; & amp ; t & lt ; 2 \end { align } $ here 's what that region looks like in the input space . next , we consider all possible outputs of the function in that range . input $ ( s , t ) $ | output $ ( t^3 -st , s-t , s+t ) $ -|- $ ( 0 , 0 ) $ | $ ( 0 , 0 , 0 ) $ $ ( 1 , 0 ) $ | $ ( 1 , 1 , 1 ) $ $ ( 2 , 1 ) $ | $ ( 6 , 1 , 3 ) $ $ \quad\vdots $ | $ \quad\ ; \vdots $ okay , so we do n't literally write out all possible outputs , since , you know , that involves infinitely many things . in principle , though , our goal is to represent all those values . since the function spits out a three-coordinate output , we visualize this output in three-dimensional space . the following animation shows what it looks like as the points $ ( s , t ) $ in the parameter space move to the corresponding output $ f ( s , t ) $ in three-dimensional space : the resulting surface in three-dimensional space is called a parametric surface . warning : surfaces like these can be confused with the graphs of functions that have a two-dimensional input and a one-dimensional output since those are also drawn as surfaces in three-dimensional space . but these parametric functions have a very different flavor . they have a two-dimensional input and a three-dimensional output . notice , this means graphing them would require five dimensions ! parameterizing a surface one of the best ways to get a feel for parametric functions is to start with a surface that you want to describe , then try to find a function that will draw this as a parametric surface . this is also a necessary skill when you start learning about surface integrals later on in multivariable calculus . be warned , though , parameterizing surfaces is not easy . in the following example , we will parameterize a torus , the fancy word for the surface of a doughnut . as surfaces go , a torus is a relatively simple example , but it still takes some serious effort . example : parameterize a torus ( doughnut ) consider the surface pictured above . you can think of it as a doughnut shape , or perhaps just as the glaze on the doughnut , since we do n't care about the filling . our goal right now is to find a function with a two-dimensional input , and a three-dimensional output , such that this doughnut shape is the output . we imagine `` drawing '' the surface , although one does not simply draw a surface with a pencil and paper the way we can draw a curve . instead , our strategy will be to draw each circular slice of the torus . to see what i mean , here is a sample of those circular slices ( drawn in blue ) : i also drew a large red circle on the $ xy $ -plane running through the centers of each of these slices . this is not part of the torus , but will be a useful reference point for the end goal of drawing each blue slice . in a real problem , the radius of the red circle might be given to you , as would the radius of each blue circular slice . for now , let 's choose arbitrarily that the radius of the red circle is $ 3 $ , and the radius of each blue slice is $ 1 $ , with the understanding that choosing different values would give different toruses ( torii ? torotes ? ) . core idea : we will describe each point on the torus as the sum of two vectors : a vector $ \vec { \textbf { c } } $ from the origin to a point on the red circle . to specify which point on the red circle , we will make this a vector-valued function that depends on a parameter $ t $ . as the value of $ t $ changes , the point on the red circle described by $ \vec { \textbf { c } } ( t ) $ will change . a vector $ \vec { \textbf { d } } $ from that point on the red circle to a point on the corresponding `` slice '' of the torus . the direction this vector points will depend on which point of the red circle it is anchored to , so the value of $ \vec { \textbf { d } } $ should depend on the parameter $ t $ used to describe points of the red circle . furthermore , we will use a second parameter $ u $ to determine which part of the blue torus slice $ \vec { \textbf { d } } $ points to . this means points on the torus will each be described as a sum . $ \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) $ ( if you are unfamiliar with the tip-to-tail method of adding vectors , consider reviewing this video ) . why this strategy ? the idea here is that we do n't immediately know how to define points on a torus , but we do know how to define circles . since the big red circle is flat on the $ xy $ -plane , and has radius $ 3 $ , we can parameterize it as follows : $ \begin { align } \quad \vec { \textbf { c } } ( t ) = 3 \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] = 3 \cos ( t ) \hat { \textbf { i } } + 3 \sin ( t ) \hat { \textbf { j } } + 0 \hat { \textbf { k } } \end { align } $ now , the vector-valued function $ \vec { \textbf { d } } ( u , t ) $ should also describe a circle , but it 's a bit trickier . the ( blue ) circular slice of the torus we want $ \vec { \textbf { d } } ( u , t ) $ to draw is at an angle . how do you draw a circle which is sitting at an angle in three-dimensional space ? well , let 's start from what we know . we know that in two dimensions , a unit circle centered at the origin can be described with the parametric function $ \begin { align } \quad g ( u ) = \left [ \begin { array } { a } \cos ( u ) \ \sin ( u ) \end { array } \right ] = \cos ( u ) \hat { \textbf { i } } + \sin ( u ) \hat { \textbf { j } } \end { align } $ for our desired blue circular slice , we do something similar , but we exchange $ \hat { \textbf { i } } $ and $ \hat { \textbf { j } } $ for different unit vectors . take a look at this picture : instead of the `` sideways '' direction being $ \hat { \textbf { i } } $ , the unit vector in the $ x $ -direction , we think of it as the unit vector pointing away from the origin , which we 'll call $ \hat { \textbf { v } } $ . actually , since that direction might depend on where we start , $ \hat { \textbf { v } } $ should be a vector-valued function dependent on the parameter $ t $ , so we write it as $ \hat { \textbf { v } } ( t ) $ . similarly the `` upward '' direction is no longer $ \hat { \textbf { j } } $ , but $ \hat { \textbf { k } } $ , the unit vector in the $ z $ -direction . therefore , the parameterization of the circular slice should look something like this : $ \begin { align } \quad \vec { \textbf { d } } ( u , t ) = \cos ( u ) \hat { \textbf { v } } ( t ) + \sin ( u ) \hat { \textbf { k } } \end { align } $ this of course leaves us with a question : what is the formula for $ \hat { \textbf { v } } ( t ) $ ? looking at the picture , the direction away from the origin is also described by $ \vec { \textbf { c } } ( t ) $ , so the formula for $ \hat { \textbf { v } } ( t ) $ should be the same as that for $ \vec { \textbf { c } } ( t ) $ , but scaled down to be a unit vector . $ \begin { align } \quad \vec { \textbf { c } } ( t ) = 3 & amp ; \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] \quad \leftarrow \small { \gray { \text { not a unit vector } } } \ \downarrow & amp ; \ \hat { \textbf { v } } ( t ) = & amp ; \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] \quad \leftarrow \small { \gray { \text { unit vector } } } \end { align } $ this means our full expression for $ \vec { \textbf { d } } ( u , t ) $ is $ \begin { align } \quad \vec { \textbf { d } } ( u , t ) & amp ; = \cos ( u ) \hat { \textbf { v } } ( t ) + \sin ( u ) \hat { \textbf { k } } \ & amp ; = \cos ( u ) \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] + \sin ( u ) \left [ \begin { array } { c } 0 \ 0 \ 1 \end { array } \right ] = \left [ \begin { array } { c } \cos ( u ) \cos ( t ) \ \cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] \end { align } $ bring it on home remember , the whole reason we defined $ \vec { \textbf { d } } ( u , t ) $ and $ \vec { \textbf { c } } ( t ) $ was to describe each point on the torus as $ \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) $ . putting this together , we have the following vector-valued two-parameter function : $ \begin { align } \quad \vec { f } ( u , t ) & amp ; = \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) \ & amp ; = 3\left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] + \left [ \begin { array } { c } \cos ( u ) \cos ( t ) \ \cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] \ & amp ; = \bluee { \left [ \begin { array } { c } 3\cos ( t ) + \cos ( u ) \cos ( t ) \ 3\sin ( t ) +\cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] } \end { align } $ as $ u $ ranges from $ 0 $ to $ 2\pi $ , the output of this function $ \vec { f } ( u , t ) $ will trace one of the blue slices , and as $ t $ ranges from $ 0 $ to $ 2\pi $ , the slices themselves will trace out the entire torus . here 's what it might look like if we take the points from the parameter space where $ 0 \le u \le 2\pi $ and $ 0 \le t \le 2\pi $ , and watch them move to the output of our function $ \vec { f } ( u , t ) $ : summary you can visualize a function with a two-dimensional input and a three-dimensional output by plotting all the output points corresponding to some region of the input space . this results in a surface , known as a parametric surface . the process of going the other way around , starting with a surface in space and trying to find a function that `` draws '' this surface , is known as parameterizing the surface . in general , this is a tricky thing to do .
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actually , since that direction might depend on where we start , $ \hat { \textbf { v } } $ should be a vector-valued function dependent on the parameter $ t $ , so we write it as $ \hat { \textbf { v } } ( t ) $ . similarly the `` upward '' direction is no longer $ \hat { \textbf { j } } $ , but $ \hat { \textbf { k } } $ , the unit vector in the $ z $ -direction . therefore , the parameterization of the circular slice should look something like this : $ \begin { align } \quad \vec { \textbf { d } } ( u , t ) = \cos ( u ) \hat { \textbf { v } } ( t ) + \sin ( u ) \hat { \textbf { k } } \end { align } $ this of course leaves us with a question : what is the formula for $ \hat { \textbf { v } } ( t ) $ ?
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why this article says that the unit vector of 3cos ( t ) i+3sin ( t ) j+0k is just cos ( t ) i+sin ( t ) j , it would not be the vector itself over its magnitude ?
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background multivariable functions parametric functions , one parameter what we 're building to you can visualize a function with a two-dimensional input and a three-dimensional output by plotting all the output points corresponding to some region of the input space . this results in a surface , known as a parametric surface . the process of going the other way around , starting with a surface in space and trying to find a function that `` draws '' this surface , is known as parameterizing the surface . in general , this is a tricky thing to do . quick review of one-parameter functions in the last article , i talked about visualizing functions with a one-dimensional input and a two-dimensional output . for example : $ \displaystyle f ( t ) = \left [ \begin { array } { c } t\cos ( t ) \ \sin ( t ) \end { array } \right ] $ i talked about how because the output space has more dimensions than the input space , you can get a good feel for the function just by seeing which points in the output space are `` hit '' by the function as the input $ t $ ranges over some set of values . when a function is interpreted this way , it is known as a parametric function , and its input $ t $ is called the parameter two parameters we can do something very similar with functions that have a two-dimensional input and a three-dimensional output . $ \displaystyle f ( s , t ) = \left [ \begin { array } { c } t^3 - st \ s-t \ s+t \end { array } \right ] $ both input coordinates $ s $ and $ t $ will be known as the parameters , and you are about to see how this function draws a surface in three-dimensional space . the first step to representing a function like this is to specify a range for the input , such as $ \begin { align } \quad 0 & lt ; & amp ; s & lt ; 3 \ -2 & lt ; & amp ; t & lt ; 2 \end { align } $ here 's what that region looks like in the input space . next , we consider all possible outputs of the function in that range . input $ ( s , t ) $ | output $ ( t^3 -st , s-t , s+t ) $ -|- $ ( 0 , 0 ) $ | $ ( 0 , 0 , 0 ) $ $ ( 1 , 0 ) $ | $ ( 1 , 1 , 1 ) $ $ ( 2 , 1 ) $ | $ ( 6 , 1 , 3 ) $ $ \quad\vdots $ | $ \quad\ ; \vdots $ okay , so we do n't literally write out all possible outputs , since , you know , that involves infinitely many things . in principle , though , our goal is to represent all those values . since the function spits out a three-coordinate output , we visualize this output in three-dimensional space . the following animation shows what it looks like as the points $ ( s , t ) $ in the parameter space move to the corresponding output $ f ( s , t ) $ in three-dimensional space : the resulting surface in three-dimensional space is called a parametric surface . warning : surfaces like these can be confused with the graphs of functions that have a two-dimensional input and a one-dimensional output since those are also drawn as surfaces in three-dimensional space . but these parametric functions have a very different flavor . they have a two-dimensional input and a three-dimensional output . notice , this means graphing them would require five dimensions ! parameterizing a surface one of the best ways to get a feel for parametric functions is to start with a surface that you want to describe , then try to find a function that will draw this as a parametric surface . this is also a necessary skill when you start learning about surface integrals later on in multivariable calculus . be warned , though , parameterizing surfaces is not easy . in the following example , we will parameterize a torus , the fancy word for the surface of a doughnut . as surfaces go , a torus is a relatively simple example , but it still takes some serious effort . example : parameterize a torus ( doughnut ) consider the surface pictured above . you can think of it as a doughnut shape , or perhaps just as the glaze on the doughnut , since we do n't care about the filling . our goal right now is to find a function with a two-dimensional input , and a three-dimensional output , such that this doughnut shape is the output . we imagine `` drawing '' the surface , although one does not simply draw a surface with a pencil and paper the way we can draw a curve . instead , our strategy will be to draw each circular slice of the torus . to see what i mean , here is a sample of those circular slices ( drawn in blue ) : i also drew a large red circle on the $ xy $ -plane running through the centers of each of these slices . this is not part of the torus , but will be a useful reference point for the end goal of drawing each blue slice . in a real problem , the radius of the red circle might be given to you , as would the radius of each blue circular slice . for now , let 's choose arbitrarily that the radius of the red circle is $ 3 $ , and the radius of each blue slice is $ 1 $ , with the understanding that choosing different values would give different toruses ( torii ? torotes ? ) . core idea : we will describe each point on the torus as the sum of two vectors : a vector $ \vec { \textbf { c } } $ from the origin to a point on the red circle . to specify which point on the red circle , we will make this a vector-valued function that depends on a parameter $ t $ . as the value of $ t $ changes , the point on the red circle described by $ \vec { \textbf { c } } ( t ) $ will change . a vector $ \vec { \textbf { d } } $ from that point on the red circle to a point on the corresponding `` slice '' of the torus . the direction this vector points will depend on which point of the red circle it is anchored to , so the value of $ \vec { \textbf { d } } $ should depend on the parameter $ t $ used to describe points of the red circle . furthermore , we will use a second parameter $ u $ to determine which part of the blue torus slice $ \vec { \textbf { d } } $ points to . this means points on the torus will each be described as a sum . $ \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) $ ( if you are unfamiliar with the tip-to-tail method of adding vectors , consider reviewing this video ) . why this strategy ? the idea here is that we do n't immediately know how to define points on a torus , but we do know how to define circles . since the big red circle is flat on the $ xy $ -plane , and has radius $ 3 $ , we can parameterize it as follows : $ \begin { align } \quad \vec { \textbf { c } } ( t ) = 3 \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] = 3 \cos ( t ) \hat { \textbf { i } } + 3 \sin ( t ) \hat { \textbf { j } } + 0 \hat { \textbf { k } } \end { align } $ now , the vector-valued function $ \vec { \textbf { d } } ( u , t ) $ should also describe a circle , but it 's a bit trickier . the ( blue ) circular slice of the torus we want $ \vec { \textbf { d } } ( u , t ) $ to draw is at an angle . how do you draw a circle which is sitting at an angle in three-dimensional space ? well , let 's start from what we know . we know that in two dimensions , a unit circle centered at the origin can be described with the parametric function $ \begin { align } \quad g ( u ) = \left [ \begin { array } { a } \cos ( u ) \ \sin ( u ) \end { array } \right ] = \cos ( u ) \hat { \textbf { i } } + \sin ( u ) \hat { \textbf { j } } \end { align } $ for our desired blue circular slice , we do something similar , but we exchange $ \hat { \textbf { i } } $ and $ \hat { \textbf { j } } $ for different unit vectors . take a look at this picture : instead of the `` sideways '' direction being $ \hat { \textbf { i } } $ , the unit vector in the $ x $ -direction , we think of it as the unit vector pointing away from the origin , which we 'll call $ \hat { \textbf { v } } $ . actually , since that direction might depend on where we start , $ \hat { \textbf { v } } $ should be a vector-valued function dependent on the parameter $ t $ , so we write it as $ \hat { \textbf { v } } ( t ) $ . similarly the `` upward '' direction is no longer $ \hat { \textbf { j } } $ , but $ \hat { \textbf { k } } $ , the unit vector in the $ z $ -direction . therefore , the parameterization of the circular slice should look something like this : $ \begin { align } \quad \vec { \textbf { d } } ( u , t ) = \cos ( u ) \hat { \textbf { v } } ( t ) + \sin ( u ) \hat { \textbf { k } } \end { align } $ this of course leaves us with a question : what is the formula for $ \hat { \textbf { v } } ( t ) $ ? looking at the picture , the direction away from the origin is also described by $ \vec { \textbf { c } } ( t ) $ , so the formula for $ \hat { \textbf { v } } ( t ) $ should be the same as that for $ \vec { \textbf { c } } ( t ) $ , but scaled down to be a unit vector . $ \begin { align } \quad \vec { \textbf { c } } ( t ) = 3 & amp ; \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] \quad \leftarrow \small { \gray { \text { not a unit vector } } } \ \downarrow & amp ; \ \hat { \textbf { v } } ( t ) = & amp ; \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] \quad \leftarrow \small { \gray { \text { unit vector } } } \end { align } $ this means our full expression for $ \vec { \textbf { d } } ( u , t ) $ is $ \begin { align } \quad \vec { \textbf { d } } ( u , t ) & amp ; = \cos ( u ) \hat { \textbf { v } } ( t ) + \sin ( u ) \hat { \textbf { k } } \ & amp ; = \cos ( u ) \left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] + \sin ( u ) \left [ \begin { array } { c } 0 \ 0 \ 1 \end { array } \right ] = \left [ \begin { array } { c } \cos ( u ) \cos ( t ) \ \cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] \end { align } $ bring it on home remember , the whole reason we defined $ \vec { \textbf { d } } ( u , t ) $ and $ \vec { \textbf { c } } ( t ) $ was to describe each point on the torus as $ \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) $ . putting this together , we have the following vector-valued two-parameter function : $ \begin { align } \quad \vec { f } ( u , t ) & amp ; = \vec { \textbf { c } } ( t ) + \vec { \textbf { d } } ( u , t ) \ & amp ; = 3\left [ \begin { array } { c } \cos ( t ) \ \sin ( t ) \ 0 \end { array } \right ] + \left [ \begin { array } { c } \cos ( u ) \cos ( t ) \ \cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] \ & amp ; = \bluee { \left [ \begin { array } { c } 3\cos ( t ) + \cos ( u ) \cos ( t ) \ 3\sin ( t ) +\cos ( u ) \sin ( t ) \ \sin ( u ) \end { array } \right ] } \end { align } $ as $ u $ ranges from $ 0 $ to $ 2\pi $ , the output of this function $ \vec { f } ( u , t ) $ will trace one of the blue slices , and as $ t $ ranges from $ 0 $ to $ 2\pi $ , the slices themselves will trace out the entire torus . here 's what it might look like if we take the points from the parameter space where $ 0 \le u \le 2\pi $ and $ 0 \le t \le 2\pi $ , and watch them move to the output of our function $ \vec { f } ( u , t ) $ : summary you can visualize a function with a two-dimensional input and a three-dimensional output by plotting all the output points corresponding to some region of the input space . this results in a surface , known as a parametric surface . the process of going the other way around , starting with a surface in space and trying to find a function that `` draws '' this surface , is known as parameterizing the surface . in general , this is a tricky thing to do .
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$ \displaystyle f ( s , t ) = \left [ \begin { array } { c } t^3 - st \ s-t \ s+t \end { array } \right ] $ both input coordinates $ s $ and $ t $ will be known as the parameters , and you are about to see how this function draws a surface in three-dimensional space . the first step to representing a function like this is to specify a range for the input , such as $ \begin { align } \quad 0 & lt ; & amp ; s & lt ; 3 \ -2 & lt ; & amp ; t & lt ; 2 \end { align } $ here 's what that region looks like in the input space . next , we consider all possible outputs of the function in that range .
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why is the input range 0 < s < 3 and -2 < t < 2 in the first example ?
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we ’ re going to guess that most of you can name a few things you ’ d rather be doing than cracking into sat prep . studying for the sat - or for any subject - can be both frustrating and difficult , not to mention boring . thankfully , there are some things you can do to make your study time more efficient ( waste less time ! ) and effective ( do the right things ! ) . we asked high school students how they prepare for tests and demanding assignments . here are some of their recommendations , which we hope you ’ ll find useful and inspiring as you embark on your own sat practice plan . how to… understand the material teach and learn : use online resources. “ looking up videos on websites such as youtube and bozeman science helps me to study and understand concepts more profoundly so that i am completely comfortable with the material. ” – eillen use your own words . “ i find that the best way to study content is to read about it while taking notes , and then explaining the notes and information to myself as if i were teaching it , in order to understand it in my own words. ” – fariha try teaching the concepts to someone else . `` if you ca n't find someone to teach it to , teach it to a blank wall or stuffed animal – anything works . you can also film yourself teaching these concepts ( film yourself discussing what happened during a certain war , film yourself on a whiteboard solving a math problem and explaining why you 're doing everything you 're doing , etc ) and then to go back and watch it and see where you 're hesitant ( hesitating means you 're not as confident on that part ) or if you 've left out any information. ” – elyse make learning interactive. “ to make studying enjoyable , it 's important to make learning interactive . for instance , i use screen recording apps to create my own lectures , and post them on a private website so that i can watch them later to review . also , the process of recording these videos helps reinforce concepts in a way that suits my learning style . this allows me to participate in your learning process , which makes studying more fun in general. ” – gaeun talk it out . “ if you happen to be studying at school , use your friends as resources . talk out the steps to solving a math problem , explain a scientific concept , or discuss the structure of your essay . verbalizing concepts and ideas can help solidify them and make them easier to complete and understand later on . studying with a friend may even be beneficial for both of you ! ” – emily put everything on paper : be prepared . “ always have a paper and pencil with you to note down anything that comes to your mind. ” – rushi take notes . “ i find it crucial to take good notes . i always make my own personalized textbook for every subject , which includes important ideas i 'm learning , personal comments about them ( especially in subjects such as literature or history ) , and the mistakes i 've made in the past on tests or on essays . it 's also helpful if you write down mnemonic devices in the margins , such as a memorable quote made by someone in class , to memorize dates or formulas . – gaeun use technology . `` i like to digitalize the content of my notes by saving them in icloud , google docs , or evernote . this enables you to access your notes on other devices or in other locations . a few weeks before big tests , good , organized notes turn out to be very useful. ” – gaeun get creative : have fun . “ i have learned to have fun with my notes . i personalize them so that they help me . sometimes , instead of scribbling everything my teacher says during a lecture i will draw pictures that i associate with the topic ( probably the only time doodling in class is considered acceptable ) . for example , we were recently reviewing a cardiovascular unit in anatomy . instead of writing out the steps of blood flow through the heart , i drew a basic diagram of the heart and labeled arrows directing the path of blood flow . instead of having some abstract concept in my head of the path of blood , i could more deeply understand the concept with the visual aid. ” – eillen use color . “ i 'm the type of person who learns by writing things out , but writing things over and over again can be a bit dull . i am able to make it through these study sessions by using different colored thick markers ; it makes studying seem more like an arts and crafts project plus writing in markers is much easier than writing in pencil . i spend time making diagrams and coming up with fun mnemonic devices for things that need to be memorized. ” – tiffany try recording yourself . “ read your notes into a voice recording app , and replay them as you do something else . the combination of reading something and hearing it again is more effective than reading alone. ” – erin challenge yourself . “ for people who are competitive like me , challenging myself and making studying a competition is what keeps me engaged . i like to use quizlet flashcard sets because they have an option where the cards are shuffled and turned into a matching game . in my opinion , making studying a competition or game is the best way to keep me engaged. ” – kennedy how to… stay focused remove temptation : turn off your phone . “ how do we avoid falling into the trap of social media and internet distractions ? i 'll tell you one way to do it . take that phone , turn it off , put it in another room , and detach yourself for a few hours - it wo n't kill you. ” – eillen commit to study time . “ to stay focused , i turn off all my devices , and just allocate some time to just study . i tell myself that i ca n't stop until the time is up , regardless of if i know the material or not. ” – neil take breaks when you need them. “ during the weekends , i know that i wo n't get any work done during the day , because i wo n't have the motivation to finish my work until it becomes urgent . therefore , i forget about my homework and allow myself to have fun and relax . this prepares me for the time when i finally have to get to work and finish my homework and study. ” – alexis keep things fun : get in the zone with music . “ music without vocals helps me focus while i study . i explore other genres that i do n't get much exposure to - flamenco music , indian classical , erhu solos , minimalist , etc . it 's a good opportunity to expand your cultural horizon while providing a non-distracting drone of sound. ” – eric treat yourself. “ to make studying fun , i take breaks every 20 minutes . i 'll get up and stretch , treat myself to a snack and a youtube video , or clean up my room ( because organization is everything ! ) . i 'll continually re-evaluate my progress , just so i know that i 'm going somewhere with all the time and energy that i 've invested. ” – heeju mix it up. “ taking lots of breaks while studying to do something completely different - going out for a jog , playing some piano , listening to some music - helps make learning more efficient by allowing for a memory refresh and refocus. ” – jody how to… study for the sat in particular understand what you know and what you don ’ t know : review your mistakes thoroughly. “ the most important thing you can possibly do when preparing for the sat is to review your mistakes . it does n't matter if the mistake you made was silly or if it happened because you were genuinely confused . all mistakes should be recorded and reviewed . for the math and reading sections , write down or cut and paste the questions you missed , and make sure to solve the problems once more after initially writing down the correct solution . a strategy that helped me was to create my own reading sections out of a set of hard passages and questions. ” – gaeun create a routine . “ the most helpful thing i did for the sat was to practice a single section a day and go into each problem i missed in detail - then i did full length practice tests every other week in the last two months before the actual exam. ” – aneesh practice at the edge of your ability. “ try to find the hardest problems , and solve those . if you ca n't do them , take some steps back. ” – rushi skip the hard ones. “ if you do n't know a question , skip it . sometimes , things do n't `` click '' , and that 's alright . just keep going and go back to the question later . most of the time you 'll realize that it was actually super easy ! ” – clayton memorize the formulas . “ know your math formulas ! there are some that are super helpful and key which are easy to find online and easy to memorize . those can be super helpful. ” – alexis keep on practicing ! practice ! the sat is something you can study for and see marked improvement . anyone can improve with practice in anything , and that 's certainly true of the sat. ” – aneesh practice ! “ the best way to study for the sat is to practice . practice , practice , practice . oh yeah , and did i mention to practice ? practice taking parts of the exam in a testing environment . time yourself . check your answers once you 're finished . if you got a question wrong , understand why you got it wrong . the practice wo n't do you any good if you do n't understand what you did incorrectly . find out what you 're good at and not so good at . work on your trouble areas . i guarantee that with practice , your score will improve. ” – eillen take full practice tests ! “ full practice tests are also invaluable . taking at least one or two before the actual test helps you gain some sense of what it 's like to sit for 4 hours taking the sat . it is essential to time yourself strictly and accurately when taking these tests. ” – gaeun take advantage of all resources : take ownership - ask questions . “ the sat is a big deal , but you are the one who ultimately decides how much you want to freak out about it . talk to people you know who have experience with taking the sat , and make sure that you are well acquainted with how the process works . speak with teachers or counselors at school who have knowledge about the process if you have questions. ” – emily take initiative - go for it ! “ practice tests and questions are available online , in bookstores , and at school . if you put in effort , there really is no need to pay for an expensive sat preparation class . the materials you need to do well on the test are all available to you , you just need to have the initiative to take advantage of them. ” – eillen do it your way : take control . “ you have to study at your pace . do n't be scared if a peer knows or understands more than you . you might just have to learn that thing another way . think about how you like to learn and that 's how you should study. ” – ryann know what works for you and stick to it . `` if you learn best in groups , hold a group study session and exchange solutions with friends . if you learn best by writing things down , write down the math formulas in fun colors . begin studying by doing what is best for you and then work your way up to taking official tests in the environment that you are expected to take the test in . that way , you can learn what your favorite strategy for taking the sat is and perfect it. ” – tiffany take the sat when it best fits your schedule . `` many students take it in october of their junior years , but that time may be too hectic for you , or you might not feel prepared enough . any time from sophomore year to early senior year is okay ; just make sure your scores will be ready by the time college apps come around. ” – shruthi free study technology quizlet : flashcards and memorization games studyblue : custom flashcards and open-source flashcard library evernote : note-taking and note management program google calendar : online calendar and task management system kahoot : online quiz and learning game program grammar girl : podcasts and articles on general grammar skills and topics youtube : videos on every topic under the sun
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most of the time you 'll realize that it was actually super easy ! ” – clayton memorize the formulas . “ know your math formulas ! there are some that are super helpful and key which are easy to find online and easy to memorize .
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continue studying my high school math and start practicing sat math in my senior year of school ?
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we ’ re going to guess that most of you can name a few things you ’ d rather be doing than cracking into sat prep . studying for the sat - or for any subject - can be both frustrating and difficult , not to mention boring . thankfully , there are some things you can do to make your study time more efficient ( waste less time ! ) and effective ( do the right things ! ) . we asked high school students how they prepare for tests and demanding assignments . here are some of their recommendations , which we hope you ’ ll find useful and inspiring as you embark on your own sat practice plan . how to… understand the material teach and learn : use online resources. “ looking up videos on websites such as youtube and bozeman science helps me to study and understand concepts more profoundly so that i am completely comfortable with the material. ” – eillen use your own words . “ i find that the best way to study content is to read about it while taking notes , and then explaining the notes and information to myself as if i were teaching it , in order to understand it in my own words. ” – fariha try teaching the concepts to someone else . `` if you ca n't find someone to teach it to , teach it to a blank wall or stuffed animal – anything works . you can also film yourself teaching these concepts ( film yourself discussing what happened during a certain war , film yourself on a whiteboard solving a math problem and explaining why you 're doing everything you 're doing , etc ) and then to go back and watch it and see where you 're hesitant ( hesitating means you 're not as confident on that part ) or if you 've left out any information. ” – elyse make learning interactive. “ to make studying enjoyable , it 's important to make learning interactive . for instance , i use screen recording apps to create my own lectures , and post them on a private website so that i can watch them later to review . also , the process of recording these videos helps reinforce concepts in a way that suits my learning style . this allows me to participate in your learning process , which makes studying more fun in general. ” – gaeun talk it out . “ if you happen to be studying at school , use your friends as resources . talk out the steps to solving a math problem , explain a scientific concept , or discuss the structure of your essay . verbalizing concepts and ideas can help solidify them and make them easier to complete and understand later on . studying with a friend may even be beneficial for both of you ! ” – emily put everything on paper : be prepared . “ always have a paper and pencil with you to note down anything that comes to your mind. ” – rushi take notes . “ i find it crucial to take good notes . i always make my own personalized textbook for every subject , which includes important ideas i 'm learning , personal comments about them ( especially in subjects such as literature or history ) , and the mistakes i 've made in the past on tests or on essays . it 's also helpful if you write down mnemonic devices in the margins , such as a memorable quote made by someone in class , to memorize dates or formulas . – gaeun use technology . `` i like to digitalize the content of my notes by saving them in icloud , google docs , or evernote . this enables you to access your notes on other devices or in other locations . a few weeks before big tests , good , organized notes turn out to be very useful. ” – gaeun get creative : have fun . “ i have learned to have fun with my notes . i personalize them so that they help me . sometimes , instead of scribbling everything my teacher says during a lecture i will draw pictures that i associate with the topic ( probably the only time doodling in class is considered acceptable ) . for example , we were recently reviewing a cardiovascular unit in anatomy . instead of writing out the steps of blood flow through the heart , i drew a basic diagram of the heart and labeled arrows directing the path of blood flow . instead of having some abstract concept in my head of the path of blood , i could more deeply understand the concept with the visual aid. ” – eillen use color . “ i 'm the type of person who learns by writing things out , but writing things over and over again can be a bit dull . i am able to make it through these study sessions by using different colored thick markers ; it makes studying seem more like an arts and crafts project plus writing in markers is much easier than writing in pencil . i spend time making diagrams and coming up with fun mnemonic devices for things that need to be memorized. ” – tiffany try recording yourself . “ read your notes into a voice recording app , and replay them as you do something else . the combination of reading something and hearing it again is more effective than reading alone. ” – erin challenge yourself . “ for people who are competitive like me , challenging myself and making studying a competition is what keeps me engaged . i like to use quizlet flashcard sets because they have an option where the cards are shuffled and turned into a matching game . in my opinion , making studying a competition or game is the best way to keep me engaged. ” – kennedy how to… stay focused remove temptation : turn off your phone . “ how do we avoid falling into the trap of social media and internet distractions ? i 'll tell you one way to do it . take that phone , turn it off , put it in another room , and detach yourself for a few hours - it wo n't kill you. ” – eillen commit to study time . “ to stay focused , i turn off all my devices , and just allocate some time to just study . i tell myself that i ca n't stop until the time is up , regardless of if i know the material or not. ” – neil take breaks when you need them. “ during the weekends , i know that i wo n't get any work done during the day , because i wo n't have the motivation to finish my work until it becomes urgent . therefore , i forget about my homework and allow myself to have fun and relax . this prepares me for the time when i finally have to get to work and finish my homework and study. ” – alexis keep things fun : get in the zone with music . “ music without vocals helps me focus while i study . i explore other genres that i do n't get much exposure to - flamenco music , indian classical , erhu solos , minimalist , etc . it 's a good opportunity to expand your cultural horizon while providing a non-distracting drone of sound. ” – eric treat yourself. “ to make studying fun , i take breaks every 20 minutes . i 'll get up and stretch , treat myself to a snack and a youtube video , or clean up my room ( because organization is everything ! ) . i 'll continually re-evaluate my progress , just so i know that i 'm going somewhere with all the time and energy that i 've invested. ” – heeju mix it up. “ taking lots of breaks while studying to do something completely different - going out for a jog , playing some piano , listening to some music - helps make learning more efficient by allowing for a memory refresh and refocus. ” – jody how to… study for the sat in particular understand what you know and what you don ’ t know : review your mistakes thoroughly. “ the most important thing you can possibly do when preparing for the sat is to review your mistakes . it does n't matter if the mistake you made was silly or if it happened because you were genuinely confused . all mistakes should be recorded and reviewed . for the math and reading sections , write down or cut and paste the questions you missed , and make sure to solve the problems once more after initially writing down the correct solution . a strategy that helped me was to create my own reading sections out of a set of hard passages and questions. ” – gaeun create a routine . “ the most helpful thing i did for the sat was to practice a single section a day and go into each problem i missed in detail - then i did full length practice tests every other week in the last two months before the actual exam. ” – aneesh practice at the edge of your ability. “ try to find the hardest problems , and solve those . if you ca n't do them , take some steps back. ” – rushi skip the hard ones. “ if you do n't know a question , skip it . sometimes , things do n't `` click '' , and that 's alright . just keep going and go back to the question later . most of the time you 'll realize that it was actually super easy ! ” – clayton memorize the formulas . “ know your math formulas ! there are some that are super helpful and key which are easy to find online and easy to memorize . those can be super helpful. ” – alexis keep on practicing ! practice ! the sat is something you can study for and see marked improvement . anyone can improve with practice in anything , and that 's certainly true of the sat. ” – aneesh practice ! “ the best way to study for the sat is to practice . practice , practice , practice . oh yeah , and did i mention to practice ? practice taking parts of the exam in a testing environment . time yourself . check your answers once you 're finished . if you got a question wrong , understand why you got it wrong . the practice wo n't do you any good if you do n't understand what you did incorrectly . find out what you 're good at and not so good at . work on your trouble areas . i guarantee that with practice , your score will improve. ” – eillen take full practice tests ! “ full practice tests are also invaluable . taking at least one or two before the actual test helps you gain some sense of what it 's like to sit for 4 hours taking the sat . it is essential to time yourself strictly and accurately when taking these tests. ” – gaeun take advantage of all resources : take ownership - ask questions . “ the sat is a big deal , but you are the one who ultimately decides how much you want to freak out about it . talk to people you know who have experience with taking the sat , and make sure that you are well acquainted with how the process works . speak with teachers or counselors at school who have knowledge about the process if you have questions. ” – emily take initiative - go for it ! “ practice tests and questions are available online , in bookstores , and at school . if you put in effort , there really is no need to pay for an expensive sat preparation class . the materials you need to do well on the test are all available to you , you just need to have the initiative to take advantage of them. ” – eillen do it your way : take control . “ you have to study at your pace . do n't be scared if a peer knows or understands more than you . you might just have to learn that thing another way . think about how you like to learn and that 's how you should study. ” – ryann know what works for you and stick to it . `` if you learn best in groups , hold a group study session and exchange solutions with friends . if you learn best by writing things down , write down the math formulas in fun colors . begin studying by doing what is best for you and then work your way up to taking official tests in the environment that you are expected to take the test in . that way , you can learn what your favorite strategy for taking the sat is and perfect it. ” – tiffany take the sat when it best fits your schedule . `` many students take it in october of their junior years , but that time may be too hectic for you , or you might not feel prepared enough . any time from sophomore year to early senior year is okay ; just make sure your scores will be ready by the time college apps come around. ” – shruthi free study technology quizlet : flashcards and memorization games studyblue : custom flashcards and open-source flashcard library evernote : note-taking and note management program google calendar : online calendar and task management system kahoot : online quiz and learning game program grammar girl : podcasts and articles on general grammar skills and topics youtube : videos on every topic under the sun
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anyone can improve with practice in anything , and that 's certainly true of the sat. ” – aneesh practice ! “ the best way to study for the sat is to practice . practice , practice , practice .
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what is best for sat , self study or coaching study ?
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we ’ re going to guess that most of you can name a few things you ’ d rather be doing than cracking into sat prep . studying for the sat - or for any subject - can be both frustrating and difficult , not to mention boring . thankfully , there are some things you can do to make your study time more efficient ( waste less time ! ) and effective ( do the right things ! ) . we asked high school students how they prepare for tests and demanding assignments . here are some of their recommendations , which we hope you ’ ll find useful and inspiring as you embark on your own sat practice plan . how to… understand the material teach and learn : use online resources. “ looking up videos on websites such as youtube and bozeman science helps me to study and understand concepts more profoundly so that i am completely comfortable with the material. ” – eillen use your own words . “ i find that the best way to study content is to read about it while taking notes , and then explaining the notes and information to myself as if i were teaching it , in order to understand it in my own words. ” – fariha try teaching the concepts to someone else . `` if you ca n't find someone to teach it to , teach it to a blank wall or stuffed animal – anything works . you can also film yourself teaching these concepts ( film yourself discussing what happened during a certain war , film yourself on a whiteboard solving a math problem and explaining why you 're doing everything you 're doing , etc ) and then to go back and watch it and see where you 're hesitant ( hesitating means you 're not as confident on that part ) or if you 've left out any information. ” – elyse make learning interactive. “ to make studying enjoyable , it 's important to make learning interactive . for instance , i use screen recording apps to create my own lectures , and post them on a private website so that i can watch them later to review . also , the process of recording these videos helps reinforce concepts in a way that suits my learning style . this allows me to participate in your learning process , which makes studying more fun in general. ” – gaeun talk it out . “ if you happen to be studying at school , use your friends as resources . talk out the steps to solving a math problem , explain a scientific concept , or discuss the structure of your essay . verbalizing concepts and ideas can help solidify them and make them easier to complete and understand later on . studying with a friend may even be beneficial for both of you ! ” – emily put everything on paper : be prepared . “ always have a paper and pencil with you to note down anything that comes to your mind. ” – rushi take notes . “ i find it crucial to take good notes . i always make my own personalized textbook for every subject , which includes important ideas i 'm learning , personal comments about them ( especially in subjects such as literature or history ) , and the mistakes i 've made in the past on tests or on essays . it 's also helpful if you write down mnemonic devices in the margins , such as a memorable quote made by someone in class , to memorize dates or formulas . – gaeun use technology . `` i like to digitalize the content of my notes by saving them in icloud , google docs , or evernote . this enables you to access your notes on other devices or in other locations . a few weeks before big tests , good , organized notes turn out to be very useful. ” – gaeun get creative : have fun . “ i have learned to have fun with my notes . i personalize them so that they help me . sometimes , instead of scribbling everything my teacher says during a lecture i will draw pictures that i associate with the topic ( probably the only time doodling in class is considered acceptable ) . for example , we were recently reviewing a cardiovascular unit in anatomy . instead of writing out the steps of blood flow through the heart , i drew a basic diagram of the heart and labeled arrows directing the path of blood flow . instead of having some abstract concept in my head of the path of blood , i could more deeply understand the concept with the visual aid. ” – eillen use color . “ i 'm the type of person who learns by writing things out , but writing things over and over again can be a bit dull . i am able to make it through these study sessions by using different colored thick markers ; it makes studying seem more like an arts and crafts project plus writing in markers is much easier than writing in pencil . i spend time making diagrams and coming up with fun mnemonic devices for things that need to be memorized. ” – tiffany try recording yourself . “ read your notes into a voice recording app , and replay them as you do something else . the combination of reading something and hearing it again is more effective than reading alone. ” – erin challenge yourself . “ for people who are competitive like me , challenging myself and making studying a competition is what keeps me engaged . i like to use quizlet flashcard sets because they have an option where the cards are shuffled and turned into a matching game . in my opinion , making studying a competition or game is the best way to keep me engaged. ” – kennedy how to… stay focused remove temptation : turn off your phone . “ how do we avoid falling into the trap of social media and internet distractions ? i 'll tell you one way to do it . take that phone , turn it off , put it in another room , and detach yourself for a few hours - it wo n't kill you. ” – eillen commit to study time . “ to stay focused , i turn off all my devices , and just allocate some time to just study . i tell myself that i ca n't stop until the time is up , regardless of if i know the material or not. ” – neil take breaks when you need them. “ during the weekends , i know that i wo n't get any work done during the day , because i wo n't have the motivation to finish my work until it becomes urgent . therefore , i forget about my homework and allow myself to have fun and relax . this prepares me for the time when i finally have to get to work and finish my homework and study. ” – alexis keep things fun : get in the zone with music . “ music without vocals helps me focus while i study . i explore other genres that i do n't get much exposure to - flamenco music , indian classical , erhu solos , minimalist , etc . it 's a good opportunity to expand your cultural horizon while providing a non-distracting drone of sound. ” – eric treat yourself. “ to make studying fun , i take breaks every 20 minutes . i 'll get up and stretch , treat myself to a snack and a youtube video , or clean up my room ( because organization is everything ! ) . i 'll continually re-evaluate my progress , just so i know that i 'm going somewhere with all the time and energy that i 've invested. ” – heeju mix it up. “ taking lots of breaks while studying to do something completely different - going out for a jog , playing some piano , listening to some music - helps make learning more efficient by allowing for a memory refresh and refocus. ” – jody how to… study for the sat in particular understand what you know and what you don ’ t know : review your mistakes thoroughly. “ the most important thing you can possibly do when preparing for the sat is to review your mistakes . it does n't matter if the mistake you made was silly or if it happened because you were genuinely confused . all mistakes should be recorded and reviewed . for the math and reading sections , write down or cut and paste the questions you missed , and make sure to solve the problems once more after initially writing down the correct solution . a strategy that helped me was to create my own reading sections out of a set of hard passages and questions. ” – gaeun create a routine . “ the most helpful thing i did for the sat was to practice a single section a day and go into each problem i missed in detail - then i did full length practice tests every other week in the last two months before the actual exam. ” – aneesh practice at the edge of your ability. “ try to find the hardest problems , and solve those . if you ca n't do them , take some steps back. ” – rushi skip the hard ones. “ if you do n't know a question , skip it . sometimes , things do n't `` click '' , and that 's alright . just keep going and go back to the question later . most of the time you 'll realize that it was actually super easy ! ” – clayton memorize the formulas . “ know your math formulas ! there are some that are super helpful and key which are easy to find online and easy to memorize . those can be super helpful. ” – alexis keep on practicing ! practice ! the sat is something you can study for and see marked improvement . anyone can improve with practice in anything , and that 's certainly true of the sat. ” – aneesh practice ! “ the best way to study for the sat is to practice . practice , practice , practice . oh yeah , and did i mention to practice ? practice taking parts of the exam in a testing environment . time yourself . check your answers once you 're finished . if you got a question wrong , understand why you got it wrong . the practice wo n't do you any good if you do n't understand what you did incorrectly . find out what you 're good at and not so good at . work on your trouble areas . i guarantee that with practice , your score will improve. ” – eillen take full practice tests ! “ full practice tests are also invaluable . taking at least one or two before the actual test helps you gain some sense of what it 's like to sit for 4 hours taking the sat . it is essential to time yourself strictly and accurately when taking these tests. ” – gaeun take advantage of all resources : take ownership - ask questions . “ the sat is a big deal , but you are the one who ultimately decides how much you want to freak out about it . talk to people you know who have experience with taking the sat , and make sure that you are well acquainted with how the process works . speak with teachers or counselors at school who have knowledge about the process if you have questions. ” – emily take initiative - go for it ! “ practice tests and questions are available online , in bookstores , and at school . if you put in effort , there really is no need to pay for an expensive sat preparation class . the materials you need to do well on the test are all available to you , you just need to have the initiative to take advantage of them. ” – eillen do it your way : take control . “ you have to study at your pace . do n't be scared if a peer knows or understands more than you . you might just have to learn that thing another way . think about how you like to learn and that 's how you should study. ” – ryann know what works for you and stick to it . `` if you learn best in groups , hold a group study session and exchange solutions with friends . if you learn best by writing things down , write down the math formulas in fun colors . begin studying by doing what is best for you and then work your way up to taking official tests in the environment that you are expected to take the test in . that way , you can learn what your favorite strategy for taking the sat is and perfect it. ” – tiffany take the sat when it best fits your schedule . `` many students take it in october of their junior years , but that time may be too hectic for you , or you might not feel prepared enough . any time from sophomore year to early senior year is okay ; just make sure your scores will be ready by the time college apps come around. ” – shruthi free study technology quizlet : flashcards and memorization games studyblue : custom flashcards and open-source flashcard library evernote : note-taking and note management program google calendar : online calendar and task management system kahoot : online quiz and learning game program grammar girl : podcasts and articles on general grammar skills and topics youtube : videos on every topic under the sun
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most of the time you 'll realize that it was actually super easy ! ” – clayton memorize the formulas . “ know your math formulas ! there are some that are super helpful and key which are easy to find online and easy to memorize .
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where can you find all the formulas required for sat math ?
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we ’ re going to guess that most of you can name a few things you ’ d rather be doing than cracking into sat prep . studying for the sat - or for any subject - can be both frustrating and difficult , not to mention boring . thankfully , there are some things you can do to make your study time more efficient ( waste less time ! ) and effective ( do the right things ! ) . we asked high school students how they prepare for tests and demanding assignments . here are some of their recommendations , which we hope you ’ ll find useful and inspiring as you embark on your own sat practice plan . how to… understand the material teach and learn : use online resources. “ looking up videos on websites such as youtube and bozeman science helps me to study and understand concepts more profoundly so that i am completely comfortable with the material. ” – eillen use your own words . “ i find that the best way to study content is to read about it while taking notes , and then explaining the notes and information to myself as if i were teaching it , in order to understand it in my own words. ” – fariha try teaching the concepts to someone else . `` if you ca n't find someone to teach it to , teach it to a blank wall or stuffed animal – anything works . you can also film yourself teaching these concepts ( film yourself discussing what happened during a certain war , film yourself on a whiteboard solving a math problem and explaining why you 're doing everything you 're doing , etc ) and then to go back and watch it and see where you 're hesitant ( hesitating means you 're not as confident on that part ) or if you 've left out any information. ” – elyse make learning interactive. “ to make studying enjoyable , it 's important to make learning interactive . for instance , i use screen recording apps to create my own lectures , and post them on a private website so that i can watch them later to review . also , the process of recording these videos helps reinforce concepts in a way that suits my learning style . this allows me to participate in your learning process , which makes studying more fun in general. ” – gaeun talk it out . “ if you happen to be studying at school , use your friends as resources . talk out the steps to solving a math problem , explain a scientific concept , or discuss the structure of your essay . verbalizing concepts and ideas can help solidify them and make them easier to complete and understand later on . studying with a friend may even be beneficial for both of you ! ” – emily put everything on paper : be prepared . “ always have a paper and pencil with you to note down anything that comes to your mind. ” – rushi take notes . “ i find it crucial to take good notes . i always make my own personalized textbook for every subject , which includes important ideas i 'm learning , personal comments about them ( especially in subjects such as literature or history ) , and the mistakes i 've made in the past on tests or on essays . it 's also helpful if you write down mnemonic devices in the margins , such as a memorable quote made by someone in class , to memorize dates or formulas . – gaeun use technology . `` i like to digitalize the content of my notes by saving them in icloud , google docs , or evernote . this enables you to access your notes on other devices or in other locations . a few weeks before big tests , good , organized notes turn out to be very useful. ” – gaeun get creative : have fun . “ i have learned to have fun with my notes . i personalize them so that they help me . sometimes , instead of scribbling everything my teacher says during a lecture i will draw pictures that i associate with the topic ( probably the only time doodling in class is considered acceptable ) . for example , we were recently reviewing a cardiovascular unit in anatomy . instead of writing out the steps of blood flow through the heart , i drew a basic diagram of the heart and labeled arrows directing the path of blood flow . instead of having some abstract concept in my head of the path of blood , i could more deeply understand the concept with the visual aid. ” – eillen use color . “ i 'm the type of person who learns by writing things out , but writing things over and over again can be a bit dull . i am able to make it through these study sessions by using different colored thick markers ; it makes studying seem more like an arts and crafts project plus writing in markers is much easier than writing in pencil . i spend time making diagrams and coming up with fun mnemonic devices for things that need to be memorized. ” – tiffany try recording yourself . “ read your notes into a voice recording app , and replay them as you do something else . the combination of reading something and hearing it again is more effective than reading alone. ” – erin challenge yourself . “ for people who are competitive like me , challenging myself and making studying a competition is what keeps me engaged . i like to use quizlet flashcard sets because they have an option where the cards are shuffled and turned into a matching game . in my opinion , making studying a competition or game is the best way to keep me engaged. ” – kennedy how to… stay focused remove temptation : turn off your phone . “ how do we avoid falling into the trap of social media and internet distractions ? i 'll tell you one way to do it . take that phone , turn it off , put it in another room , and detach yourself for a few hours - it wo n't kill you. ” – eillen commit to study time . “ to stay focused , i turn off all my devices , and just allocate some time to just study . i tell myself that i ca n't stop until the time is up , regardless of if i know the material or not. ” – neil take breaks when you need them. “ during the weekends , i know that i wo n't get any work done during the day , because i wo n't have the motivation to finish my work until it becomes urgent . therefore , i forget about my homework and allow myself to have fun and relax . this prepares me for the time when i finally have to get to work and finish my homework and study. ” – alexis keep things fun : get in the zone with music . “ music without vocals helps me focus while i study . i explore other genres that i do n't get much exposure to - flamenco music , indian classical , erhu solos , minimalist , etc . it 's a good opportunity to expand your cultural horizon while providing a non-distracting drone of sound. ” – eric treat yourself. “ to make studying fun , i take breaks every 20 minutes . i 'll get up and stretch , treat myself to a snack and a youtube video , or clean up my room ( because organization is everything ! ) . i 'll continually re-evaluate my progress , just so i know that i 'm going somewhere with all the time and energy that i 've invested. ” – heeju mix it up. “ taking lots of breaks while studying to do something completely different - going out for a jog , playing some piano , listening to some music - helps make learning more efficient by allowing for a memory refresh and refocus. ” – jody how to… study for the sat in particular understand what you know and what you don ’ t know : review your mistakes thoroughly. “ the most important thing you can possibly do when preparing for the sat is to review your mistakes . it does n't matter if the mistake you made was silly or if it happened because you were genuinely confused . all mistakes should be recorded and reviewed . for the math and reading sections , write down or cut and paste the questions you missed , and make sure to solve the problems once more after initially writing down the correct solution . a strategy that helped me was to create my own reading sections out of a set of hard passages and questions. ” – gaeun create a routine . “ the most helpful thing i did for the sat was to practice a single section a day and go into each problem i missed in detail - then i did full length practice tests every other week in the last two months before the actual exam. ” – aneesh practice at the edge of your ability. “ try to find the hardest problems , and solve those . if you ca n't do them , take some steps back. ” – rushi skip the hard ones. “ if you do n't know a question , skip it . sometimes , things do n't `` click '' , and that 's alright . just keep going and go back to the question later . most of the time you 'll realize that it was actually super easy ! ” – clayton memorize the formulas . “ know your math formulas ! there are some that are super helpful and key which are easy to find online and easy to memorize . those can be super helpful. ” – alexis keep on practicing ! practice ! the sat is something you can study for and see marked improvement . anyone can improve with practice in anything , and that 's certainly true of the sat. ” – aneesh practice ! “ the best way to study for the sat is to practice . practice , practice , practice . oh yeah , and did i mention to practice ? practice taking parts of the exam in a testing environment . time yourself . check your answers once you 're finished . if you got a question wrong , understand why you got it wrong . the practice wo n't do you any good if you do n't understand what you did incorrectly . find out what you 're good at and not so good at . work on your trouble areas . i guarantee that with practice , your score will improve. ” – eillen take full practice tests ! “ full practice tests are also invaluable . taking at least one or two before the actual test helps you gain some sense of what it 's like to sit for 4 hours taking the sat . it is essential to time yourself strictly and accurately when taking these tests. ” – gaeun take advantage of all resources : take ownership - ask questions . “ the sat is a big deal , but you are the one who ultimately decides how much you want to freak out about it . talk to people you know who have experience with taking the sat , and make sure that you are well acquainted with how the process works . speak with teachers or counselors at school who have knowledge about the process if you have questions. ” – emily take initiative - go for it ! “ practice tests and questions are available online , in bookstores , and at school . if you put in effort , there really is no need to pay for an expensive sat preparation class . the materials you need to do well on the test are all available to you , you just need to have the initiative to take advantage of them. ” – eillen do it your way : take control . “ you have to study at your pace . do n't be scared if a peer knows or understands more than you . you might just have to learn that thing another way . think about how you like to learn and that 's how you should study. ” – ryann know what works for you and stick to it . `` if you learn best in groups , hold a group study session and exchange solutions with friends . if you learn best by writing things down , write down the math formulas in fun colors . begin studying by doing what is best for you and then work your way up to taking official tests in the environment that you are expected to take the test in . that way , you can learn what your favorite strategy for taking the sat is and perfect it. ” – tiffany take the sat when it best fits your schedule . `` many students take it in october of their junior years , but that time may be too hectic for you , or you might not feel prepared enough . any time from sophomore year to early senior year is okay ; just make sure your scores will be ready by the time college apps come around. ” – shruthi free study technology quizlet : flashcards and memorization games studyblue : custom flashcards and open-source flashcard library evernote : note-taking and note management program google calendar : online calendar and task management system kahoot : online quiz and learning game program grammar girl : podcasts and articles on general grammar skills and topics youtube : videos on every topic under the sun
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anyone can improve with practice in anything , and that 's certainly true of the sat. ” – aneesh practice ! “ the best way to study for the sat is to practice . practice , practice , practice .
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what 's the best thing i can do for getting a good grip over the sat ?
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we ’ re going to guess that most of you can name a few things you ’ d rather be doing than cracking into sat prep . studying for the sat - or for any subject - can be both frustrating and difficult , not to mention boring . thankfully , there are some things you can do to make your study time more efficient ( waste less time ! ) and effective ( do the right things ! ) . we asked high school students how they prepare for tests and demanding assignments . here are some of their recommendations , which we hope you ’ ll find useful and inspiring as you embark on your own sat practice plan . how to… understand the material teach and learn : use online resources. “ looking up videos on websites such as youtube and bozeman science helps me to study and understand concepts more profoundly so that i am completely comfortable with the material. ” – eillen use your own words . “ i find that the best way to study content is to read about it while taking notes , and then explaining the notes and information to myself as if i were teaching it , in order to understand it in my own words. ” – fariha try teaching the concepts to someone else . `` if you ca n't find someone to teach it to , teach it to a blank wall or stuffed animal – anything works . you can also film yourself teaching these concepts ( film yourself discussing what happened during a certain war , film yourself on a whiteboard solving a math problem and explaining why you 're doing everything you 're doing , etc ) and then to go back and watch it and see where you 're hesitant ( hesitating means you 're not as confident on that part ) or if you 've left out any information. ” – elyse make learning interactive. “ to make studying enjoyable , it 's important to make learning interactive . for instance , i use screen recording apps to create my own lectures , and post them on a private website so that i can watch them later to review . also , the process of recording these videos helps reinforce concepts in a way that suits my learning style . this allows me to participate in your learning process , which makes studying more fun in general. ” – gaeun talk it out . “ if you happen to be studying at school , use your friends as resources . talk out the steps to solving a math problem , explain a scientific concept , or discuss the structure of your essay . verbalizing concepts and ideas can help solidify them and make them easier to complete and understand later on . studying with a friend may even be beneficial for both of you ! ” – emily put everything on paper : be prepared . “ always have a paper and pencil with you to note down anything that comes to your mind. ” – rushi take notes . “ i find it crucial to take good notes . i always make my own personalized textbook for every subject , which includes important ideas i 'm learning , personal comments about them ( especially in subjects such as literature or history ) , and the mistakes i 've made in the past on tests or on essays . it 's also helpful if you write down mnemonic devices in the margins , such as a memorable quote made by someone in class , to memorize dates or formulas . – gaeun use technology . `` i like to digitalize the content of my notes by saving them in icloud , google docs , or evernote . this enables you to access your notes on other devices or in other locations . a few weeks before big tests , good , organized notes turn out to be very useful. ” – gaeun get creative : have fun . “ i have learned to have fun with my notes . i personalize them so that they help me . sometimes , instead of scribbling everything my teacher says during a lecture i will draw pictures that i associate with the topic ( probably the only time doodling in class is considered acceptable ) . for example , we were recently reviewing a cardiovascular unit in anatomy . instead of writing out the steps of blood flow through the heart , i drew a basic diagram of the heart and labeled arrows directing the path of blood flow . instead of having some abstract concept in my head of the path of blood , i could more deeply understand the concept with the visual aid. ” – eillen use color . “ i 'm the type of person who learns by writing things out , but writing things over and over again can be a bit dull . i am able to make it through these study sessions by using different colored thick markers ; it makes studying seem more like an arts and crafts project plus writing in markers is much easier than writing in pencil . i spend time making diagrams and coming up with fun mnemonic devices for things that need to be memorized. ” – tiffany try recording yourself . “ read your notes into a voice recording app , and replay them as you do something else . the combination of reading something and hearing it again is more effective than reading alone. ” – erin challenge yourself . “ for people who are competitive like me , challenging myself and making studying a competition is what keeps me engaged . i like to use quizlet flashcard sets because they have an option where the cards are shuffled and turned into a matching game . in my opinion , making studying a competition or game is the best way to keep me engaged. ” – kennedy how to… stay focused remove temptation : turn off your phone . “ how do we avoid falling into the trap of social media and internet distractions ? i 'll tell you one way to do it . take that phone , turn it off , put it in another room , and detach yourself for a few hours - it wo n't kill you. ” – eillen commit to study time . “ to stay focused , i turn off all my devices , and just allocate some time to just study . i tell myself that i ca n't stop until the time is up , regardless of if i know the material or not. ” – neil take breaks when you need them. “ during the weekends , i know that i wo n't get any work done during the day , because i wo n't have the motivation to finish my work until it becomes urgent . therefore , i forget about my homework and allow myself to have fun and relax . this prepares me for the time when i finally have to get to work and finish my homework and study. ” – alexis keep things fun : get in the zone with music . “ music without vocals helps me focus while i study . i explore other genres that i do n't get much exposure to - flamenco music , indian classical , erhu solos , minimalist , etc . it 's a good opportunity to expand your cultural horizon while providing a non-distracting drone of sound. ” – eric treat yourself. “ to make studying fun , i take breaks every 20 minutes . i 'll get up and stretch , treat myself to a snack and a youtube video , or clean up my room ( because organization is everything ! ) . i 'll continually re-evaluate my progress , just so i know that i 'm going somewhere with all the time and energy that i 've invested. ” – heeju mix it up. “ taking lots of breaks while studying to do something completely different - going out for a jog , playing some piano , listening to some music - helps make learning more efficient by allowing for a memory refresh and refocus. ” – jody how to… study for the sat in particular understand what you know and what you don ’ t know : review your mistakes thoroughly. “ the most important thing you can possibly do when preparing for the sat is to review your mistakes . it does n't matter if the mistake you made was silly or if it happened because you were genuinely confused . all mistakes should be recorded and reviewed . for the math and reading sections , write down or cut and paste the questions you missed , and make sure to solve the problems once more after initially writing down the correct solution . a strategy that helped me was to create my own reading sections out of a set of hard passages and questions. ” – gaeun create a routine . “ the most helpful thing i did for the sat was to practice a single section a day and go into each problem i missed in detail - then i did full length practice tests every other week in the last two months before the actual exam. ” – aneesh practice at the edge of your ability. “ try to find the hardest problems , and solve those . if you ca n't do them , take some steps back. ” – rushi skip the hard ones. “ if you do n't know a question , skip it . sometimes , things do n't `` click '' , and that 's alright . just keep going and go back to the question later . most of the time you 'll realize that it was actually super easy ! ” – clayton memorize the formulas . “ know your math formulas ! there are some that are super helpful and key which are easy to find online and easy to memorize . those can be super helpful. ” – alexis keep on practicing ! practice ! the sat is something you can study for and see marked improvement . anyone can improve with practice in anything , and that 's certainly true of the sat. ” – aneesh practice ! “ the best way to study for the sat is to practice . practice , practice , practice . oh yeah , and did i mention to practice ? practice taking parts of the exam in a testing environment . time yourself . check your answers once you 're finished . if you got a question wrong , understand why you got it wrong . the practice wo n't do you any good if you do n't understand what you did incorrectly . find out what you 're good at and not so good at . work on your trouble areas . i guarantee that with practice , your score will improve. ” – eillen take full practice tests ! “ full practice tests are also invaluable . taking at least one or two before the actual test helps you gain some sense of what it 's like to sit for 4 hours taking the sat . it is essential to time yourself strictly and accurately when taking these tests. ” – gaeun take advantage of all resources : take ownership - ask questions . “ the sat is a big deal , but you are the one who ultimately decides how much you want to freak out about it . talk to people you know who have experience with taking the sat , and make sure that you are well acquainted with how the process works . speak with teachers or counselors at school who have knowledge about the process if you have questions. ” – emily take initiative - go for it ! “ practice tests and questions are available online , in bookstores , and at school . if you put in effort , there really is no need to pay for an expensive sat preparation class . the materials you need to do well on the test are all available to you , you just need to have the initiative to take advantage of them. ” – eillen do it your way : take control . “ you have to study at your pace . do n't be scared if a peer knows or understands more than you . you might just have to learn that thing another way . think about how you like to learn and that 's how you should study. ” – ryann know what works for you and stick to it . `` if you learn best in groups , hold a group study session and exchange solutions with friends . if you learn best by writing things down , write down the math formulas in fun colors . begin studying by doing what is best for you and then work your way up to taking official tests in the environment that you are expected to take the test in . that way , you can learn what your favorite strategy for taking the sat is and perfect it. ” – tiffany take the sat when it best fits your schedule . `` many students take it in october of their junior years , but that time may be too hectic for you , or you might not feel prepared enough . any time from sophomore year to early senior year is okay ; just make sure your scores will be ready by the time college apps come around. ” – shruthi free study technology quizlet : flashcards and memorization games studyblue : custom flashcards and open-source flashcard library evernote : note-taking and note management program google calendar : online calendar and task management system kahoot : online quiz and learning game program grammar girl : podcasts and articles on general grammar skills and topics youtube : videos on every topic under the sun
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speak with teachers or counselors at school who have knowledge about the process if you have questions. ” – emily take initiative - go for it ! “ practice tests and questions are available online , in bookstores , and at school . if you put in effort , there really is no need to pay for an expensive sat preparation class . the materials you need to do well on the test are all available to you , you just need to have the initiative to take advantage of them. ” – eillen do it your way : take control . “ you have to study at your pace .
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how many words do i need to write in the essay part ?
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we ’ re going to guess that most of you can name a few things you ’ d rather be doing than cracking into sat prep . studying for the sat - or for any subject - can be both frustrating and difficult , not to mention boring . thankfully , there are some things you can do to make your study time more efficient ( waste less time ! ) and effective ( do the right things ! ) . we asked high school students how they prepare for tests and demanding assignments . here are some of their recommendations , which we hope you ’ ll find useful and inspiring as you embark on your own sat practice plan . how to… understand the material teach and learn : use online resources. “ looking up videos on websites such as youtube and bozeman science helps me to study and understand concepts more profoundly so that i am completely comfortable with the material. ” – eillen use your own words . “ i find that the best way to study content is to read about it while taking notes , and then explaining the notes and information to myself as if i were teaching it , in order to understand it in my own words. ” – fariha try teaching the concepts to someone else . `` if you ca n't find someone to teach it to , teach it to a blank wall or stuffed animal – anything works . you can also film yourself teaching these concepts ( film yourself discussing what happened during a certain war , film yourself on a whiteboard solving a math problem and explaining why you 're doing everything you 're doing , etc ) and then to go back and watch it and see where you 're hesitant ( hesitating means you 're not as confident on that part ) or if you 've left out any information. ” – elyse make learning interactive. “ to make studying enjoyable , it 's important to make learning interactive . for instance , i use screen recording apps to create my own lectures , and post them on a private website so that i can watch them later to review . also , the process of recording these videos helps reinforce concepts in a way that suits my learning style . this allows me to participate in your learning process , which makes studying more fun in general. ” – gaeun talk it out . “ if you happen to be studying at school , use your friends as resources . talk out the steps to solving a math problem , explain a scientific concept , or discuss the structure of your essay . verbalizing concepts and ideas can help solidify them and make them easier to complete and understand later on . studying with a friend may even be beneficial for both of you ! ” – emily put everything on paper : be prepared . “ always have a paper and pencil with you to note down anything that comes to your mind. ” – rushi take notes . “ i find it crucial to take good notes . i always make my own personalized textbook for every subject , which includes important ideas i 'm learning , personal comments about them ( especially in subjects such as literature or history ) , and the mistakes i 've made in the past on tests or on essays . it 's also helpful if you write down mnemonic devices in the margins , such as a memorable quote made by someone in class , to memorize dates or formulas . – gaeun use technology . `` i like to digitalize the content of my notes by saving them in icloud , google docs , or evernote . this enables you to access your notes on other devices or in other locations . a few weeks before big tests , good , organized notes turn out to be very useful. ” – gaeun get creative : have fun . “ i have learned to have fun with my notes . i personalize them so that they help me . sometimes , instead of scribbling everything my teacher says during a lecture i will draw pictures that i associate with the topic ( probably the only time doodling in class is considered acceptable ) . for example , we were recently reviewing a cardiovascular unit in anatomy . instead of writing out the steps of blood flow through the heart , i drew a basic diagram of the heart and labeled arrows directing the path of blood flow . instead of having some abstract concept in my head of the path of blood , i could more deeply understand the concept with the visual aid. ” – eillen use color . “ i 'm the type of person who learns by writing things out , but writing things over and over again can be a bit dull . i am able to make it through these study sessions by using different colored thick markers ; it makes studying seem more like an arts and crafts project plus writing in markers is much easier than writing in pencil . i spend time making diagrams and coming up with fun mnemonic devices for things that need to be memorized. ” – tiffany try recording yourself . “ read your notes into a voice recording app , and replay them as you do something else . the combination of reading something and hearing it again is more effective than reading alone. ” – erin challenge yourself . “ for people who are competitive like me , challenging myself and making studying a competition is what keeps me engaged . i like to use quizlet flashcard sets because they have an option where the cards are shuffled and turned into a matching game . in my opinion , making studying a competition or game is the best way to keep me engaged. ” – kennedy how to… stay focused remove temptation : turn off your phone . “ how do we avoid falling into the trap of social media and internet distractions ? i 'll tell you one way to do it . take that phone , turn it off , put it in another room , and detach yourself for a few hours - it wo n't kill you. ” – eillen commit to study time . “ to stay focused , i turn off all my devices , and just allocate some time to just study . i tell myself that i ca n't stop until the time is up , regardless of if i know the material or not. ” – neil take breaks when you need them. “ during the weekends , i know that i wo n't get any work done during the day , because i wo n't have the motivation to finish my work until it becomes urgent . therefore , i forget about my homework and allow myself to have fun and relax . this prepares me for the time when i finally have to get to work and finish my homework and study. ” – alexis keep things fun : get in the zone with music . “ music without vocals helps me focus while i study . i explore other genres that i do n't get much exposure to - flamenco music , indian classical , erhu solos , minimalist , etc . it 's a good opportunity to expand your cultural horizon while providing a non-distracting drone of sound. ” – eric treat yourself. “ to make studying fun , i take breaks every 20 minutes . i 'll get up and stretch , treat myself to a snack and a youtube video , or clean up my room ( because organization is everything ! ) . i 'll continually re-evaluate my progress , just so i know that i 'm going somewhere with all the time and energy that i 've invested. ” – heeju mix it up. “ taking lots of breaks while studying to do something completely different - going out for a jog , playing some piano , listening to some music - helps make learning more efficient by allowing for a memory refresh and refocus. ” – jody how to… study for the sat in particular understand what you know and what you don ’ t know : review your mistakes thoroughly. “ the most important thing you can possibly do when preparing for the sat is to review your mistakes . it does n't matter if the mistake you made was silly or if it happened because you were genuinely confused . all mistakes should be recorded and reviewed . for the math and reading sections , write down or cut and paste the questions you missed , and make sure to solve the problems once more after initially writing down the correct solution . a strategy that helped me was to create my own reading sections out of a set of hard passages and questions. ” – gaeun create a routine . “ the most helpful thing i did for the sat was to practice a single section a day and go into each problem i missed in detail - then i did full length practice tests every other week in the last two months before the actual exam. ” – aneesh practice at the edge of your ability. “ try to find the hardest problems , and solve those . if you ca n't do them , take some steps back. ” – rushi skip the hard ones. “ if you do n't know a question , skip it . sometimes , things do n't `` click '' , and that 's alright . just keep going and go back to the question later . most of the time you 'll realize that it was actually super easy ! ” – clayton memorize the formulas . “ know your math formulas ! there are some that are super helpful and key which are easy to find online and easy to memorize . those can be super helpful. ” – alexis keep on practicing ! practice ! the sat is something you can study for and see marked improvement . anyone can improve with practice in anything , and that 's certainly true of the sat. ” – aneesh practice ! “ the best way to study for the sat is to practice . practice , practice , practice . oh yeah , and did i mention to practice ? practice taking parts of the exam in a testing environment . time yourself . check your answers once you 're finished . if you got a question wrong , understand why you got it wrong . the practice wo n't do you any good if you do n't understand what you did incorrectly . find out what you 're good at and not so good at . work on your trouble areas . i guarantee that with practice , your score will improve. ” – eillen take full practice tests ! “ full practice tests are also invaluable . taking at least one or two before the actual test helps you gain some sense of what it 's like to sit for 4 hours taking the sat . it is essential to time yourself strictly and accurately when taking these tests. ” – gaeun take advantage of all resources : take ownership - ask questions . “ the sat is a big deal , but you are the one who ultimately decides how much you want to freak out about it . talk to people you know who have experience with taking the sat , and make sure that you are well acquainted with how the process works . speak with teachers or counselors at school who have knowledge about the process if you have questions. ” – emily take initiative - go for it ! “ practice tests and questions are available online , in bookstores , and at school . if you put in effort , there really is no need to pay for an expensive sat preparation class . the materials you need to do well on the test are all available to you , you just need to have the initiative to take advantage of them. ” – eillen do it your way : take control . “ you have to study at your pace . do n't be scared if a peer knows or understands more than you . you might just have to learn that thing another way . think about how you like to learn and that 's how you should study. ” – ryann know what works for you and stick to it . `` if you learn best in groups , hold a group study session and exchange solutions with friends . if you learn best by writing things down , write down the math formulas in fun colors . begin studying by doing what is best for you and then work your way up to taking official tests in the environment that you are expected to take the test in . that way , you can learn what your favorite strategy for taking the sat is and perfect it. ” – tiffany take the sat when it best fits your schedule . `` many students take it in october of their junior years , but that time may be too hectic for you , or you might not feel prepared enough . any time from sophomore year to early senior year is okay ; just make sure your scores will be ready by the time college apps come around. ” – shruthi free study technology quizlet : flashcards and memorization games studyblue : custom flashcards and open-source flashcard library evernote : note-taking and note management program google calendar : online calendar and task management system kahoot : online quiz and learning game program grammar girl : podcasts and articles on general grammar skills and topics youtube : videos on every topic under the sun
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begin studying by doing what is best for you and then work your way up to taking official tests in the environment that you are expected to take the test in . that way , you can learn what your favorite strategy for taking the sat is and perfect it. ” – tiffany take the sat when it best fits your schedule . `` many students take it in october of their junior years , but that time may be too hectic for you , or you might not feel prepared enough .
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what 's the difference between the old and new sat ?
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we ’ re going to guess that most of you can name a few things you ’ d rather be doing than cracking into sat prep . studying for the sat - or for any subject - can be both frustrating and difficult , not to mention boring . thankfully , there are some things you can do to make your study time more efficient ( waste less time ! ) and effective ( do the right things ! ) . we asked high school students how they prepare for tests and demanding assignments . here are some of their recommendations , which we hope you ’ ll find useful and inspiring as you embark on your own sat practice plan . how to… understand the material teach and learn : use online resources. “ looking up videos on websites such as youtube and bozeman science helps me to study and understand concepts more profoundly so that i am completely comfortable with the material. ” – eillen use your own words . “ i find that the best way to study content is to read about it while taking notes , and then explaining the notes and information to myself as if i were teaching it , in order to understand it in my own words. ” – fariha try teaching the concepts to someone else . `` if you ca n't find someone to teach it to , teach it to a blank wall or stuffed animal – anything works . you can also film yourself teaching these concepts ( film yourself discussing what happened during a certain war , film yourself on a whiteboard solving a math problem and explaining why you 're doing everything you 're doing , etc ) and then to go back and watch it and see where you 're hesitant ( hesitating means you 're not as confident on that part ) or if you 've left out any information. ” – elyse make learning interactive. “ to make studying enjoyable , it 's important to make learning interactive . for instance , i use screen recording apps to create my own lectures , and post them on a private website so that i can watch them later to review . also , the process of recording these videos helps reinforce concepts in a way that suits my learning style . this allows me to participate in your learning process , which makes studying more fun in general. ” – gaeun talk it out . “ if you happen to be studying at school , use your friends as resources . talk out the steps to solving a math problem , explain a scientific concept , or discuss the structure of your essay . verbalizing concepts and ideas can help solidify them and make them easier to complete and understand later on . studying with a friend may even be beneficial for both of you ! ” – emily put everything on paper : be prepared . “ always have a paper and pencil with you to note down anything that comes to your mind. ” – rushi take notes . “ i find it crucial to take good notes . i always make my own personalized textbook for every subject , which includes important ideas i 'm learning , personal comments about them ( especially in subjects such as literature or history ) , and the mistakes i 've made in the past on tests or on essays . it 's also helpful if you write down mnemonic devices in the margins , such as a memorable quote made by someone in class , to memorize dates or formulas . – gaeun use technology . `` i like to digitalize the content of my notes by saving them in icloud , google docs , or evernote . this enables you to access your notes on other devices or in other locations . a few weeks before big tests , good , organized notes turn out to be very useful. ” – gaeun get creative : have fun . “ i have learned to have fun with my notes . i personalize them so that they help me . sometimes , instead of scribbling everything my teacher says during a lecture i will draw pictures that i associate with the topic ( probably the only time doodling in class is considered acceptable ) . for example , we were recently reviewing a cardiovascular unit in anatomy . instead of writing out the steps of blood flow through the heart , i drew a basic diagram of the heart and labeled arrows directing the path of blood flow . instead of having some abstract concept in my head of the path of blood , i could more deeply understand the concept with the visual aid. ” – eillen use color . “ i 'm the type of person who learns by writing things out , but writing things over and over again can be a bit dull . i am able to make it through these study sessions by using different colored thick markers ; it makes studying seem more like an arts and crafts project plus writing in markers is much easier than writing in pencil . i spend time making diagrams and coming up with fun mnemonic devices for things that need to be memorized. ” – tiffany try recording yourself . “ read your notes into a voice recording app , and replay them as you do something else . the combination of reading something and hearing it again is more effective than reading alone. ” – erin challenge yourself . “ for people who are competitive like me , challenging myself and making studying a competition is what keeps me engaged . i like to use quizlet flashcard sets because they have an option where the cards are shuffled and turned into a matching game . in my opinion , making studying a competition or game is the best way to keep me engaged. ” – kennedy how to… stay focused remove temptation : turn off your phone . “ how do we avoid falling into the trap of social media and internet distractions ? i 'll tell you one way to do it . take that phone , turn it off , put it in another room , and detach yourself for a few hours - it wo n't kill you. ” – eillen commit to study time . “ to stay focused , i turn off all my devices , and just allocate some time to just study . i tell myself that i ca n't stop until the time is up , regardless of if i know the material or not. ” – neil take breaks when you need them. “ during the weekends , i know that i wo n't get any work done during the day , because i wo n't have the motivation to finish my work until it becomes urgent . therefore , i forget about my homework and allow myself to have fun and relax . this prepares me for the time when i finally have to get to work and finish my homework and study. ” – alexis keep things fun : get in the zone with music . “ music without vocals helps me focus while i study . i explore other genres that i do n't get much exposure to - flamenco music , indian classical , erhu solos , minimalist , etc . it 's a good opportunity to expand your cultural horizon while providing a non-distracting drone of sound. ” – eric treat yourself. “ to make studying fun , i take breaks every 20 minutes . i 'll get up and stretch , treat myself to a snack and a youtube video , or clean up my room ( because organization is everything ! ) . i 'll continually re-evaluate my progress , just so i know that i 'm going somewhere with all the time and energy that i 've invested. ” – heeju mix it up. “ taking lots of breaks while studying to do something completely different - going out for a jog , playing some piano , listening to some music - helps make learning more efficient by allowing for a memory refresh and refocus. ” – jody how to… study for the sat in particular understand what you know and what you don ’ t know : review your mistakes thoroughly. “ the most important thing you can possibly do when preparing for the sat is to review your mistakes . it does n't matter if the mistake you made was silly or if it happened because you were genuinely confused . all mistakes should be recorded and reviewed . for the math and reading sections , write down or cut and paste the questions you missed , and make sure to solve the problems once more after initially writing down the correct solution . a strategy that helped me was to create my own reading sections out of a set of hard passages and questions. ” – gaeun create a routine . “ the most helpful thing i did for the sat was to practice a single section a day and go into each problem i missed in detail - then i did full length practice tests every other week in the last two months before the actual exam. ” – aneesh practice at the edge of your ability. “ try to find the hardest problems , and solve those . if you ca n't do them , take some steps back. ” – rushi skip the hard ones. “ if you do n't know a question , skip it . sometimes , things do n't `` click '' , and that 's alright . just keep going and go back to the question later . most of the time you 'll realize that it was actually super easy ! ” – clayton memorize the formulas . “ know your math formulas ! there are some that are super helpful and key which are easy to find online and easy to memorize . those can be super helpful. ” – alexis keep on practicing ! practice ! the sat is something you can study for and see marked improvement . anyone can improve with practice in anything , and that 's certainly true of the sat. ” – aneesh practice ! “ the best way to study for the sat is to practice . practice , practice , practice . oh yeah , and did i mention to practice ? practice taking parts of the exam in a testing environment . time yourself . check your answers once you 're finished . if you got a question wrong , understand why you got it wrong . the practice wo n't do you any good if you do n't understand what you did incorrectly . find out what you 're good at and not so good at . work on your trouble areas . i guarantee that with practice , your score will improve. ” – eillen take full practice tests ! “ full practice tests are also invaluable . taking at least one or two before the actual test helps you gain some sense of what it 's like to sit for 4 hours taking the sat . it is essential to time yourself strictly and accurately when taking these tests. ” – gaeun take advantage of all resources : take ownership - ask questions . “ the sat is a big deal , but you are the one who ultimately decides how much you want to freak out about it . talk to people you know who have experience with taking the sat , and make sure that you are well acquainted with how the process works . speak with teachers or counselors at school who have knowledge about the process if you have questions. ” – emily take initiative - go for it ! “ practice tests and questions are available online , in bookstores , and at school . if you put in effort , there really is no need to pay for an expensive sat preparation class . the materials you need to do well on the test are all available to you , you just need to have the initiative to take advantage of them. ” – eillen do it your way : take control . “ you have to study at your pace . do n't be scared if a peer knows or understands more than you . you might just have to learn that thing another way . think about how you like to learn and that 's how you should study. ” – ryann know what works for you and stick to it . `` if you learn best in groups , hold a group study session and exchange solutions with friends . if you learn best by writing things down , write down the math formulas in fun colors . begin studying by doing what is best for you and then work your way up to taking official tests in the environment that you are expected to take the test in . that way , you can learn what your favorite strategy for taking the sat is and perfect it. ” – tiffany take the sat when it best fits your schedule . `` many students take it in october of their junior years , but that time may be too hectic for you , or you might not feel prepared enough . any time from sophomore year to early senior year is okay ; just make sure your scores will be ready by the time college apps come around. ” – shruthi free study technology quizlet : flashcards and memorization games studyblue : custom flashcards and open-source flashcard library evernote : note-taking and note management program google calendar : online calendar and task management system kahoot : online quiz and learning game program grammar girl : podcasts and articles on general grammar skills and topics youtube : videos on every topic under the sun
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begin studying by doing what is best for you and then work your way up to taking official tests in the environment that you are expected to take the test in . that way , you can learn what your favorite strategy for taking the sat is and perfect it. ” – tiffany take the sat when it best fits your schedule . `` many students take it in october of their junior years , but that time may be too hectic for you , or you might not feel prepared enough .
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what is a college ready sat score ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body .
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1 ) what is a jerk used to measure ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity .
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2 ) what are the units for jerk ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ .
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how can we calculate the jerk using only the information given by a velocity-time graph ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body .
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are there quantities like acceleration of jerk , jerk of jerk and so on ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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$ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s .
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is there something like average acceleration ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time .
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jerk 's example can be giant wheel , roller coaster ... ..right ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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$ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk .
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okay , so if the slope of the at graph goes below the x-axis , how should the areas above and below compare ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval .
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what is the difference between acceleration and velocity ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body .
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why is this type of motion called one-dimensional motion ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below .
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in example 2 , sail boat windy ride , who do we know that the initial velocity is 10m/s ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ?
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in the first graph the line is comin down so that means that the acc is decreasing so its negative right ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration .
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is it possible to find the displacement from an acceleration-time graph ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity .
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last graph how did you find vi = 10 m/s ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below .
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and , if we have a constant acceleration then we wo n't experience jerk ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below .
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in example 2 would the sailboat be going in the opposite direction at 28 m/s ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below .
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when the line on the sailboat graph became `` negative '' it means it starts moving in the opposite direction and not slowing down right ?
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what does the vertical axis represent on an acceleration graph ? the vertical axis represents the acceleration of the object . for example , if you read the value of the graph shown below at a particular time , you will get the acceleration of the object in meters per second squared for that moment . try sliding the dot horizontally on the graph below to choose different times , and see how the acceleration—abbreviated acc—changes . $ $ concept check : according to the graph above , what is the acceleration at time $ t=4\text { s } $ ? what does the slope represent on an acceleration graph ? the slope of an acceleration graph represents a quantity called the jerk . the jerk is the rate of change of the acceleration . for an acceleration graph , the slope can be found from $ \text { slope } =\dfrac { \text { rise } } { \text { run } } =\dfrac { a_2-a_1 } { t_2-t_1 } =\dfrac { \delta a } { \delta t } $ , as can be seen in the diagram below . this slope , which represents the rate of change of acceleration , is defined to be the jerk . $ \text { jerk } =\dfrac { \delta a } { \delta t } $ as strange as the name jerk sounds , it fits well with what we would call jerky motion . if you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time , the motion would feel jerky , and you would have to keep applying different amounts of force from your muscles to stabilize your body . to finish up this section , let 's visualize the jerk with the example graph shown below . try moving the dot horizontally to see what the slope—i.e. , jerk—looks like at different points in time . $ $ concept check : for the acceleration graph shown above , is the jerk positive , negative , or zero at $ t=6\text { s } $ ? what does the area represent on an acceleration graph ? the area under an acceleration graph represents the change in velocity . in other words , the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval . $ \large \text { area } =\delta v $ it might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 $ ~\dfrac { \text m } { \text s^2 } $ for a time of 9 s. if we multiply both sides of the definition of acceleration , $ a=\dfrac { \delta v } { \delta t } $ , by the change in time , $ \delta t $ , we get $ \delta v=a\delta t $ . plugging in the acceleration 4 $ ~\dfrac { \text m } { \text s^2 } $ and the time interval 9 s we can find the change in velocity : $ \delta v=a\delta t= ( 4~\dfrac { \text m } { \text s^2 } ) ( 9\text { s } ) =36\dfrac { \text m } { \text s } $ multiplying the acceleration by the time interval is equivalent to finding the area under the curve . the area under the curve is a rectangle , as seen in the diagram below . the area can be found by multiplying height times width . the height of this rectangle is 4 $ ~\dfrac { \text m } { \text s^2 } $ , and the width is 9 s. so , finding the area also gives you the change in velocity . $ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s . as she nears the finish line , the race car driver starts to accelerate . the graph shown below gives the acceleration of the race car as it starts to speed up . assume the race car had a velocity of 20 m/s at time $ t=0\text { s } $ . what is the velocity of the race car after the 8 seconds of acceleration shown in the graph ? we can find the change in velocity by finding the area under the acceleration graph . $ \delta v=\text { area } =\dfrac { 1 } { 2 } bh=\dfrac { 1 } { 2 } ( 8\text { s } ) ( 6\dfrac { \text m } { \text s^2 } ) =24\text { m/s } \quad \text { ( use the formula for area of triangle : $ \dfrac { 1 } { 2 } bh. $ ) } $ $ \delta v=24\text { m/s } \quad \text { ( calculate the change in velocity . ) } $ but this is just the change in velocity during the time interval . we need to find the final velocity . we can use the definition of the change in velocity , $ \delta v=v_f-v_i $ , to find that $ \delta v=24\text { m/s } $ $ v_f-v_i=24\text { m/s } \qquad { \text { ( plug in $ v_f-v_i $ for $ \delta v $ . ) } } $ $ v_f-20\text { m/s } =24\text { m/s } \qquad { \text { ( plug in 20 m/s for the initial velocity $ v_i $ . ) } } $ $ v_f=24\text { m/s } +20\text { m/s } \qquad { \text { ( solve for $ v_f $ . ) } } $ $ v_f=44\text { m/s } \qquad { \text { ( calculate and celebrate ! ) } } $ the final velocity of the race car was 44 m/s . example 2 : sailboat windy ride a sailboat is sailing in a straight line with a velocity of 10 m/s . then at time $ t=0\text { s } $ , a stiff wind blows causing the sailboat to accelerate as seen in the diagram below . what is the velocity of the sailboat after the wind has blown for 9 seconds ? the area under the graph will give the change in velocity . the area of the graph can be broken into a rectangle , a triangle , and a triangle , as seen in the diagram below . the blue rectangle between $ t=0\text { s } $ and $ t=3\text { s } $ is considered positive area since it is above the horizontal axis . the green triangle between $ t=3\text { s } $ and $ t=7\text { s } $ is also considered positive area since it is above the horizontal axis . the red triangle between $ t=7\text { s } $ and $ t=9\text { s } $ , however , is considered negative area since it is below the horizontal axis . we 'll add these areas together—using $ hw $ for the rectangle and $ \dfrac { 1 } { 2 } bh $ for the triangles—to get the total area between $ t=0\text { s } $ and $ t=9\text { s } $ . $ \delta v=\text { area } = ( 4\dfrac { \text m } { \text s^2 } ) ( 3\text { s } ) +\dfrac { 1 } { 2 } ( 4\text { s } ) ( 4\dfrac { \text m } { \text s^2 } ) +\dfrac { 1 } { 2 } ( 2\text { s } ) ( -2\dfrac { \text m } { \text s^2 } ) \quad \text { ( add areas of rectangle and two triangles . ) } $ $ \delta v=18\text { m/s } \quad \text { ( calculate to get total change in velocity . ) } $ but this is the change in velocity , so to find the final velocity , we 'll use the definition of change in velocity . $ v_f-v_i=18\text { m/s } \quad \text { ( use definition of change in velocity . ) } $ $ v_f=18\text { m/s } +v_i\quad \text { ( solve for the final velocity . ) } $ $ v_f=18\text { m/s } +10\text { m/s } \quad \text { ( plug in initial velocity . ) } $ $ v_f=28\text { m/s } \quad \text { ( calculate and celebrate ! ) } $ the final velocity of the sailboat is $ v_f=28\text { m/s } $ .
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$ \text { area } =4~\dfrac { \text m } { \text s^2 } \times 9\text { s } =36\dfrac { \text m } { \text s } $ the area under any acceleration graph for a certain time interval gives the change in velocity for that time interval . what do solved examples involving acceleration vs. time graphs look like ? example 1 : race car acceleration a confident race car driver is cruising at a constant velocity of 20 m/s .
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how can i tell when an acceleration vs. time graph is moving at a constant acceleration ?
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