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_{A}2 plus, this is going to be mgy_{A}. Right? That's the initial height. And this is going to be equal to 12 mv_{D}2. 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_{D}2. 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_{D}2. So when you move the one half over and you combine the two terms, this actually ends up being 988 equals v_{D}2. 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.