Solving Projectile Motion Using Energy - Video Tutorials & Practice Problems

1

example

Projectile Motion with Energy

Video duration:

4m

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Hey guys, So we've already seen how powerful and useful this energy conservation equation can be. We said that we can solve new types of problems that we've never seen before using this energy conservation, but we've also said that we can solve some old familiar problems much quicker by using energy conservation project. Our motion is a perfect example of this. So you'll see some projectile motion type problems where you have objects that are thrown or launched. They're gonna asking for heights or speeds. And the idea here is that you can sometimes, which I'll talk about in a later video here. Come back to this. In a later video. You can sometimes solve these much quicker using energy conservation. So let's go ahead and walk through this example here and I'll show you what this means. So we have a ball that's thrown from the top of a building that's 30 m. So that's basically our initial height like this and we have an initial speed of 28 m per second. So we've got some velocity or initial speed which is going to be 20 like this. And what I want to do is actually want to use energy conservation not projectile motion and motion equations to figure out the speed of the ball before it hits the ground. So let's revisit the points of projectile motion. If you will, the point where you launch it is a point where it reaches the maximum height is B. Then there is the point where it comes down and reaches the initial height from which you've thrown it which is basically like the symmetrical point. And there's also the point right before it hits the ground. That's point D. So we're trying to do is we have the initial speed which is V. A. Which is 20. And we're trying to use energy conservation to figure out what the speed right before it hits the ground. This is gonna be V. D. Like this that's the target variable. We've got changing heights and speeds. We're gonna go ahead and use our energy conservation equations. We've already got a diagram Now we're gonna write out our big equation here. So it's going to be kinetic initial plus potential initial but just to kind of be consistent with the points I've made here, I'm gonna do K. A. Plus you A plus work done by non conservative equals K. D. Plus you D. Right so now we're just gonna have to go and eliminate and expand out all the terms. So that's a good kinetic energy at a, well we have some speed at a right where you have the initial speed which is 20. We know what the launch angle is but we know that the speed is 20. So therefore there is definitely some kinetic energy here. What about gravitational potential? Well, if this heights if this building here is 30 m high and that means that your initial height here is actually 30 m above the ground, therefore you definitely have some gravitational potential energy. What about work done by non conservative? Well, remember work done by non conservative. Is any works done by you and work done by friction. We're gonna ignore air resistance. So there's nothing there. But what about the work done by you? Well, you may think that you're actually in putting some work because you're throwing this ball at 20 m per second. But what happens is our problem really starts the moment right after the ball leaves your hand. So right after it leaves your hand is actually when it's traveling from A to D. And therefore there is no work that is done by you. All right. So what about kinetic energy at point D. Well, it definitely is going to be some because we're trying to calculate the speed at point D. And then finally gravitational potential. We hear when you're at the ground your Y. D. Is equal to zero. So that means your highest equal to zero. So there is no gravitational potential. All right. So let's expand all the other terms here, we have one half, then we have mass. This is gonna be via squared plus this is going to be M. G. Y. A. Right? That's the initial heights and this is going to be equal to one half M. V. D. Square. So that we're trying to figure out V. D. Here. So we can do is we can actually cancel out the mass from each one of the terms here because it appears in all the terms on the left and right side. Now, we're just gonna go ahead and start plugging in numbers here. So I've got one half. The initial speed is 20 m/s And then I've got G. Which is 9.8 times the height, which is 30. This is gonna be one half of the D. Square. All right, So just go ahead and plugging in some numbers here. This ends up being 200 this ends up being 294. And the sense of being one half V. D. Squared. So when you move the one half over and you combine the two terms, this actually ends up being 988 equals V. D. Squared. So that means you take the square roots of 988 and you're gonna get 31.4 m/s. So notice how using energy conservation this is much quicker and much more straight forward. Uh, this actually would have been a pretty difficult problem to solve because you have this unknown launch angle to deal with. So the data angle is unknown. So you would have had to use some, you know, some tricky sort of like equation systems and stuff like that to figure out the speed here. But with energy conservation becomes much more straightforward. All right, so that's how you solve these problems. Guys, let me know if you have any questions.

2

example

Calculating Max Height with Energy Conservation

Video duration:

5m

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Alright folks, so in this problem we're giving a ball that's launched at some speed and angle and instead of using the normal projectile motion equations that we've seen to solve for the maximum height, we're actually gonna use energy conservation here. So let's get started. We're gonna have to draw a diagram and set up the problem. Let's do that. So I've got the ground level like this, I've got my ball that's launched at some speed, which is 20 I'm told that the angle that that is launched at. So let's make this a little bit bigger is 37 degrees. So this ball is going to take a parabolic path like this and then it's going to come back down to the ground again. So remember that in these projectile motion problems, we have three points of interest, we have the point A the launch point, the B is the point where it reaches its maximum height. Let's be over here and that's C. Is the point where it reaches, or it goes back down to the initial height. So we want to do is we want to calculate the height here at point B. That's the maximum height. But now we want to use energy conservation to do that. So really what we have is we're gonna have to write an energy conservation equation, that's the second step. Now, if we want to figure out the height at point B, we're gonna have to pick an interval that includes that be so the easiest one to use is gonna be the point from A to B. So we're gonna have to, or the interval from A to B. So we're gonna have to set up an energy conservation equation along this interval here. So this is gonna be K. A. Plus you A plus. Work done by non conservative forces is equal to K. B. Plus you be. So now we're just gonna go ahead and limited and expand the terms. So do we have any kinetic energy at point a. We do because we're told that the initial velocity here is 20. So there's definitely some kinetic energy. What about any potential energy? Remember that's either gravitational potential or spring energy. There's no springs obviously in this problem. And we can do is we can set this point here to be zero. It may not actually be zero. Where we're just gonna call it zero because that's sort of the lowest point in our problem. And that allows us to say that the potential energy here at point A is zero. It doesn't matter. Right? So what about this work done by non conservative? Remember let's say the work that's done by you or friction. There's no you or friction. There's none of those, none of those forces that are acting. It's only gravity. So what about kinetic energy at point B. So, remember here, at point B that the velocity of the object is still just going to is perfectly horizontal at the apex. The we know that the y velocity at the peak is going to be zero but the x velocity is not, the x velocity is gonna be whatever the velocity was at point A it's gonna be V. A. X. This is just equal to V. X. Remember the X component of velocity stays the same throughout the entire projectile motion because there's no acceleration in the X axis. So what happens is there is definitely some kinetic energy here and there's also some potential energy because we're at some height above are zero points. So that's the fourth step. Now we're just gonna go ahead and expand all the terms again and solve. So this is gonna be one half M V A squared plus actually this is gonna be equal to one half M V B squared. Um Plus and this is going to be M. G. And then Y B. So this is actually we're looking for here. One thing we can do is we can cancel the M. S. Because they appear in all the terms of our problems. So really let's go ahead and look through our variables. I know what V. A. Is. I was given that in the problem and the only other variable I have is Y B. That's the target variable. Now I have a. G. Is that's just the constant. The only other variable I need is I need to figure out what's this be, what's this VB which is really just the X velocity at the peak. So in order to do that, I can actually just go ahead and say, well remember that the X velocity is going to be the same throughout the entire problem. So if I'm told what the launch speed and angle is, I can actually figure out what this X component is. So VX is just going to be 20 times the co sign of 37 and that's going to be 16. So that means that this velocity here at the peak is just 16. Not sorry, not squared is just gonna be 16 but it's just going purely to the rights like this, it's lost all of its y velocity but it's still moving with 16 to the right, Alright, so now we can go ahead and plug in. So we have one half, this is gonna be 20 squared equals one half. This is gonna be 16 squared and then plus and this is just gonna be 9.8 times Y B. Alright, so again, we just have to solve for that, we can go ahead and rearrange basically what you're gonna get here. Um Well actually 11 last thing we can do is we can actually multiply this whole entire equation by two. This just kind of gets rid of the one half that makes our equations a little bit easier. So this is just gonna be 20 squared equals 16 squared plus two times 9.8 times yb. All I've done is I've just multiplied times two throughout the entire equation. It just gets rid of the one half. All right, So now all we do is just bring the 16 square to the other side and then we just have to divide by this two times 9.8. What you're gonna get here is 20 squared minus 16 squared, divided by two times 9.8. And this will equal your maximum height. If you go ahead and work the sandwich you're gonna get is a height of 7.35 meters. And that's your final answer. Alright? That's maximum high using energy conservation. Let me know if you have any questions.

3

Problem

Problem

You are practicing jumping as far as you can. In one attempt, you run and leave the floor with 7 m/s directed at an unknown angle. What maximum height do you reach if your speed at that point is 5 m/s? Ignore air resistance.

A

1.22 m

B

1.73 m

C

2.44 m

D

0.67 m

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