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Projectile Motion with Energy

Patrick Ford
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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.