Solving Projectile Motion Using Energy - Online Tutor, Practice Problems & Exam Prep

Projectile Motion with Energy

Guys, we've already seen how powerful and useful the 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. Projectile motion is a perfect example of this. You'll see some projectile motion type problems where you have objects that are thrown or launched. They're going to be asking for heights or speeds, and the idea here is that you can sometimes, which I'll talk about in a later video, come back to this in a later video, you can sometimes solve these much quicker using energy conservation. 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 meters. So that's basically our initial height like this. And we have an initial speed of 28 meters per second. So we've got some velocity or initial speed, which is going to be 20 like this. What I want to do is actually use energy conservation, not projectile motion and motion equations, to figure out the speed of the ball before it hits the ground.

Let's revisit the points of projectile motion, if you will. The point where you launch it is A, the point where it reaches the maximum height is B, then there's 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 what 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 the speed right before it hits the ground. This is going to be v_D like this. That's the target variable. We've got changing heights and speeds. We're going to go ahead and use our energy conservation equation. So we've already got our diagram. Now, we're going to 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 going to do k_A plus u_A plus work done by non-conservative equals k_D plus u_D. Right?

Now we're just going to have to go and eliminate and expand out all the terms. Let's look at kinetic energy at A. Well, we have some speed at A. Right? We 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 building here is 30 meters high, then that means that your initial height here is actually 30 meters 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 going to ignore air resistance, so there's nothing there. What about the work done by you? Well, you may think that you're actually inputting some work because you're throwing this ball at 20 meters 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.

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. Well, here when you're at the ground, your y_D is equal to 0. So that means your height is equal to 0, so there is no gravitational potential. Right. So let's expand out all the other terms here. We have 12, then we have mass. This is going to be v_A2 plus, this is going to be mgy_A. Right? That's the initial height. And this is going to be equal to 12 mv_D2. So we're trying to figure out v_D here. So what 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 going to go ahead and start plugging in numbers here. So I've got 12, the initial speed is 20 meters per second, and then I've got g which is 9.8 times the height which is 30. This is going to be 12 v_D2. Alright? So just go ahead and plugging in some numbers here. This ends up being 200, and this ends up being 294 and this ends up being 12 v_D2. So when you move the one half over and you combine the two terms, this actually ends up being 988 equals v_D2. So that means you take the square roots of 988, and you're going to get 31.4 meters per second. So notice how using energy conservation, this is much quicker and much more straightforward. This actually would have been a pretty difficult problem to solve because you have this unknown launch angle to deal with. So the theta 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, it becomes much more straightforward. Alright? So that's how you solve these climate problems, guys. Let me know if you have any questions.

Calculating Max Height with Energy Conservation

Alright, folks. So in this problem we're given 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 going to use energy conservation here. So let's get started. We're going to 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 it is launched at, so let me 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 3 points of interest. We have point A, the launch point. Point B is the point where it reaches its maximum height, let's B over here. And point C is where it goes back down to the initial height. So what we want to do is we want to calculate the height here at point B. That's the maximum height. Now we want to use energy conservation to do that. So, really what we have is we're going to 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 going to have to pick an interval that includes that B. So the easiest one to use is going to be the point from A to B. So we're going to have to or the interval from A to B. So we're going to have to set up an energy conservation equation along this interval here. So this is going to be K_a + U_a + W_{\text{nc}} = K_b + U_b. So now we're just going to go ahead and limit it 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 it is either gravitational potential or spring energy. There are no springs obviously in this problem. And what we can do is we can set this point here to be 0. It might not actually be 0, but we're just going to call it 0 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 0. It doesn't matter. Right? So what about this work done by non-conservative forces? Remember, let's say that the work done by air resistance or friction, there is no air resistance or friction, there is 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 be perfectly horizontal at the apex. We know that the y velocity at the peak is going to be 0, but the x velocity is not. The x velocity is going to be whatever the velocity was, at point A. It's going to be v_{a_x}, This is just equal to v_x. Right? Remember the x component of the 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 our zero points. So that's the 4th step. Now we're just going to go ahead and expand all the terms again and solve. So this is going to be \frac{1}{2} m v_a^2 plus actually, this is going to be equal to \frac{1}{2} mv_b^2, plus, and this is gonna be mg y_b. So this is actually what we're looking for here. One thing we can do is we can cancel out 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 g is, that's just the constant. The only other variable I need is I need to figure out what's this b, what's this v_b, 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 v_x is just going to be 20 times the cosine 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. It's just going to be 16, but it's just going purely to the right 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 \frac{1}{2}, this is going to be 20^2 equals \frac{1}{2}, this is going to be 16^2, and then plus, and this is just going to 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 going to get here, well, actually, one last thing we can do is we can actually multiply this whole entire equation by 2. This just kind of gets rid of the one-half. It makes our equations a little bit easier. So this is just going to be 20^2 equals 16^2 plus 2 \times 9.8 \times y_b. All I've done is I've just multiplied times 2 throughout the entire equation. It just gets rid of the one-half. Alright? So now all we do is just bring the 16^2 to the other side and then we just have to divide by this 2 \times 9.8. What you're going to get here is \frac{20^2 - 16^2}{2 \times 9.8}, and this will equal your maximum height. If you go ahead and work this out, what you're going to get is a height of 7.35 meters, and that's your final answer. Alright. That's maximum height using energy conservation. Let me know if you have any questions.

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