| [6.51s -> 16.56s] Hi everyone, Physics Ninja here. In the previous problem I looked at a projectile that was launched horizontally and it calculated how long it was in the air, how far it went, and its speed right before impact. |
| [16.56s -> 28.34s] I now want to consider the same projectile. I'm going to consider the same cliff in the same setup, except this time I'm going to launch it at an angle. So let's try to calculate our same parameters. So how much time is it in the air? |
| [28.34s -> 38.13s] how far does it go distance from the cliff and also what's its final velocity so we're going to use the same parameters as before my initial velocity is going to be 20 meters per second |
| [38.58s -> 51.22s] The angle here, let's give it some arbitrary angle, let's say 25 degrees. And the height of the cliff that I'm launching the projectile from is going to be 15 meters. |
| [55.73s -> 60.75s] all right so question one is how much time is it in the air so this projectile is gonna do some |
| [61.62s -> 75.20s] Some trajectory like this. It's going to be in the air quite a bit of time So in order to find this equation we're going to proceed with the same approach We did with the previous case is we're going to look at the vertical motion, right? The reason we want to look at the vertical motion |
| [75.20s -> 89.20s] If you looked at the horizontal motion, you don't know what its horizontal displacement is by the time it gets here at the bottom, but you do know the vertical displacement. So if you look at our equation describing the vertical position of this projectile, |
| [89.23s -> 96.34s] The general equation is this one. And I'll go ahead and substitute what the acceleration is. |
| [96.75s -> 106.90s] So we have something like this. y0 is my initial position. Again, I'm going to take the y equals the 0 position right here down at the bottom of the cliff. |
| [106.90s -> 118.67s] So that would make that my initial position for this problem is 15, and it's positive 15 meters. Now the next term is the initial velocity. However, it's only the initial velocity in the vertical direction. |
| [119.28s -> 130.35s] So if you have some general velocity here at some angle, v0, the vertical component of that velocity is this one over here. |
| [131.12s -> 144.69s] This is my initial velocity in the y direction, which for this problem is pretty straightforward. Simply v0, the hypotenuse, multiplied by sine of 25 degrees. |
| [147.06s -> 159.34s] And then again, minus 1 half little gt squared. Now my final position here, my final position is when I'm going to be down at the bottom. And this is the final vertical position. So this term here is going to be 0. |
| [159.76s -> 174.11s] So again, rewriting our final equation here, we get 15 plus 20 times sine of 25. That's 8.5. Check that out in your calculator. I forgot the time here in the first expression. |
| [174.11s -> 188.34s] minus one half nine point eight t squared. So here's equation one. This is a quadratic equation in time. In order to solve this you have to remember what the quadratic equation is. |
| [188.94s -> 196.18s] remember minus b plus or minus square root of b squared minus 4ac |
| [197.20s -> 209.33s] divided by 2a. If you substitute these numbers in your calculator, so c is 15, b is 8.5, and the number a, if you group both of those together, will be 4.9. |
| [209.39s -> 220.56s] You're going to find at the time for this problem. You're going to get two solutions one of them is positive 2.82 and the other one is negative approximately 1.1 |
| [220.98s -> 228.78s] Now for all these types of problems when you're dealing with physics, we just throw away this solution. We're really only interested in the positive time. |
| [229.14s -> 236.21s] So here's the solution 2.82 seconds. That's how much time it takes me to go from the top all the way down here to the bottom |
| [239.50s -> 245.78s] Alright, our next question is how far does it go? So really all they want to know now is how far does this projectile travel? |
| [246.42s -> 260.21s] Let's call this distance x. My equation of motion for the x direction, and if you look at the previous video, simply the initial velocity, except this time it's the initial velocity in the x direction multiplied by how much time it takes. |
| [260.21s -> 272.69s] So this is a pretty straightforward problem. Again, if my original projectile was launched here at some initial velocity v0, which is 20, that would make the initial velocity in the x direction |
| [273.14s -> 284.62s] This component over here. This would be V zero X which in this case would simply be 20 and that's the adjacent side, so that's cos of |
| [285.01s -> 298.85s] 25 degrees. You substitute everything in the calculator, I think you get approximately 18.1 meters per second for that initial component. So that makes this problem pretty simple. My distance x that I travel |
| [298.85s -> 312.94s] is simply my initial velocity, which we just calculated, 18.1, multiplied by the total time. The total time was our 2.82. And this gives you approximately 51 meters. |
| [314.51s -> 316.21s] So that's it for that part. |
| [318.35s -> 331.79s] Alright, the final question here is what's the final velocity here? We've calculated both components of the initial velocity, but really what I'm asking for now is what's the final velocity when it's right down here, the instant before it hits the ground? |
| [332.08s -> 343.95s] All right, it's going at some angle over here. Well, and again, if I draw the same vector kind of down over here to make it a little bit bigger, this here has two components. |
| [344.66s -> 354.70s] All right, it's going to have an x component, a final x component, the x, and it's also going to have a final y component, which is going to be pointing down. |
| [355.06s -> 359.98s] Vy so if we're able to find both of those components we've solved the problem |
| [360.62s -> 375.28s] So let's go ahead and do that. So the x component of the velocity, again, since there's no acceleration in the x direction, it's simply the initial component of the velocity. That's easy. That's our 18.1 that we just calculated, meters per second. |
| [375.60s -> 386.22s] Now the Y direction is a little bit more complicated because in the Y direction, well initially we calculated it was 8.5, but at the top we know that it's zero and then after it turns around. |
| [387.38s -> 396.40s] So velocity in the y direction is my initial velocity in the y direction. Then there's the acceleration term, minus little g times time. |
| [396.75s -> 406.29s] So the first term is positive because it's pointing up. And we're going to take that to be positive. Acceleration is pointing down, so our second term here is negative. |
| [406.61s -> 421.58s] So all you have to do now is just substitute in the values. We get 8.5, positive, pointing up, minus 9.8, multiplied by the time. Again, the time is 2.82 seconds. That's the time we calculated in the first part. |
| [421.84s -> 432.37s] So the y component, the vertical component of the velocity for this problem ends up being negative 18.9 meters per second. |
| [432.78s -> 443.50s] So again, if you want to find the total velocity or the magnitude of the velocity, let's use Pythagorean theorem. Vx squared. |
| [443.86s -> 452.62s] Plus V y squared just plug and play at this point. We're going to get V is |
| [455.15s -> 468.98s] substitute our numbers here vx was 18.1 squared plus 18.9 squared we get a final velocity of 26.3 |
| [469.23s -> 481.58s] meters per second. All right, just to finally define the total vector, now we know what the magnitude is, but if you wanted to find the angle, again, you can use tangent here. |
| [481.90s -> 493.14s] Tangent of this angle is simply going to be Vy over Vx. So the angle theta is the inverse function. |
| [496.82s -> 509.10s] And substituting all the numbers in here, we know Vy is 18.9 18.1 for Vx and you get 46.2 degrees |
| [509.20s -> 522.83s] And again, that's the angle with respect to the horizontal, the way I've defined it here in this triangle. All right, here's one more bonus question. Typically another problem that comes up is what is this maximum height here? |
| [523.86s -> 533.94s] What is y max? It's kind of two approaches to find this one. The first one is just use our equation for the y position. |
| [534.77s -> 545.07s] All right, if I just rewrite the entire thing, minus 1 half little gt squared. All right, my initial position, well, my initial position we said was 15 meters. |
| [546.03s -> 550.74s] plus my initial velocity in the y direction was 8.5. |
| [552.37s -> 566.22s] Multiply that by the time. And minus 1 half. Little g would take 9.8. And multiply by times squared. Now I'm missing the time, right? If I know the time, I can certainly... |
| [566.22s -> 578.26s] substitute it in this equation and find the height. But I need to know how much time it takes to get to the top. However, one thing at the top that we know is that the vertical component of the velocity is zero. |
| [579.12s -> 591.09s] So if we know that, we can use our vertical velocity equation, which looks like this one, minus little g times time. This allows me to find the time at the top. |
| [591.54s -> 597.71s] So at the top we know it's zero. Let me just remind here. Just write this down |
| [598.22s -> 611.50s] Okay, I know the initial velocity in the y direction. I know little g I can solve for time. Time for this problem is simply the initial velocity divided by little g. And again, it's only the initial velocity in the vertical component. |
| [611.98s -> 622.64s] Substitute the numbers in here. We get 8.5 over 9.8. This should give us about 0.87 seconds. |
| [625.39s -> 640.14s] So all you have to do now is substitute the time back in our equation. And if you do that, what you're going to do, you're going to find that my vertical height as measured from the ground is going to be 18.7 meters. |
| [643.09s -> 656.50s] Now there's another way of doing this. If you remember, there's another kinematic equation. I'll just do it down here at the bottom. Another kinematic equation was this one. V0 squared plus 2AD. |
| [658.48s -> 667.25s] And again, in this case here, let's simplify this a little bit. Let me just actually write it as the displacement here is going to be in the y direction. |
| [667.73s -> 681.36s] And instead of writing the acceleration, let me get rid of that, I'm going to substitute little g. However, if I do that, I also have to put the direction in. So I'm going to put minus 2 times little g multiplied by the vertical displacement. |
| [681.46s -> 690.74s] Now the velocity again, it's going to be the velocity at the top and the velocity at the top now You got to be careful with this equation the velocity at the top is not zero |
| [691.09s -> 702.80s] Okay, the velocity at the top, there's still some horizontal velocity over there, right? So at the top, we still have v0x squared. |
| [703.25s -> 717.74s] This equals to the total initial velocity minus 2 times little g times delta y. You substitute in all our values here. What you're going to find is delta y. |
| [718.61s -> 731.73s] is going to be equal to 20 squared that's the initial velocity squared minus the initial x component of the velocity and divided by twice the |
| [732.53s -> 747.34s] Acceleration due to gravity. You substitute in all your values over here, you're going to find that delta y is equal to 3.7 meters. Now it looks different from this one, but remember, in this equation here, it gives us the displacement in the y direction. |
| [747.60s -> 758.96s] Delta y is really my position minus the initial position. So at the end of the day, we're going to get the exact same answer for my final position. |
| [759.47s -> 773.20s] Because the initial position here is 15 meters. Delta Y is 3.7. Therefore, the final Y has to be 18.7 meters. So kind of two different approaches to solving the same problem. Thanks for watching. |
| [773.20s -> 780.43s] If you like the channel, please subscribe. If you have any questions, don't hesitate to leave a comment or send me an email. I'd love to help you out. |
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