Energy conservation - solved example

- [Narrator] Let's use energy conservation to solve a couple of problems. Here's the first one. A cricket ball is thrown vertically up with a speed of 20 meters per second from ground. Find the speed of the ball when it has reached a height of 15 meters. Given g is 10 meters per second squared. Okay, let's draw a situation. So, we are given that a ball is thrown up from the ground with a speed of 20 meters per second. So, 20 meters per second. We're asked to find the speed of the ball when it has reached a height of 15 meters. So, as the ball goes up, we're asked to find what's the speed, what its speed, when it is somewhere over here, at a height of 15 meter. So we need to calculate what's the speed at that point. So, how do we solve this problem? Well, one way is we could use the equations of motion. We have solved problems like this before, in videos on motion and equations of motion. But, in this video, let's use energy conservation to try and solve this, okay? So, what does energy conservation say? It says, as this ball is rising up, the total energy, kinetic plus potential, that energy does not change as the ball goes from here to here. So, basically, over here, the ball has some kinetic energy, right? Now, as the ball goes up, its kinetic energy reduces, because the ball slows down, isn't it? But, that kinetic energy gets converted to potential energy, because as the ball goes higher and higher, it has more and more potential energy. So, as a result, what we see is that although the kinetic energy is reducing, potential energy is increasing, and therefore, the total energy does not change. That's the energy conservation principle. So we can now say that the total energy here should equal the total energy over here, okay? So, mathematically, we can say the total energy, which is potential plus kinetic, u is for potential, okay? So potential plus kinetic at this point, which we can call as, you know, point one, that should equal the total energy over here, we can call this as point two. And then, we can see if we can use this to try and solve our problem. So, let's see. Let's call this as point number one, and let's call this as point number two. So, we need to calculate the velocity at point number two, right? And, here's the beautiful thing about this equation. I don't care which is initial and which is final. I could have called this as one and this as two, it doesn't matter. I can also use this equation if the ball is falling down; even the direction doesn't matter over here. That's the beauty, I don't have to worry which is initial and which is final. Okay, so let's go ahead and substitute. Potential energy is mgh, because it's the potential energy to gravity, right? So over here, potential energy at one we can say it is mgh one, where h one represents the height at this point. Now, we'll substitute the values later, but let's just write it down now. So this will be mgh one plus kinetic energy at point one would be half mv one squared, where v one would be the velocity at point one. So, that should equal mgh two plus half mv two squared. Okay? Now, before we substitute, can you see that m is common everywhere? Let's pull that out. So if we take m common, then you see m cancels out. And now let's see if we have everything that we want. We know the value of g, that's given. H one, what is h one? H one is the height at this point. We know that, it's zero, right? Because I'm throwing from the ground. So I know h one is zero. I know v one, that is this number. H two? Oh, I know h two, that is the height at this point, that is 15 meters. Which means I need to calculate what v two is, and that's the only unknown over here, so I can just plug in and calculate. So you know what? Great idea to pause the video and see if you can substitute and get the answer yourself. Okay, let's substitute. So, h one is zero, therefore this will be zero. Plus half. Half v one squared, what is v one? V one is velocity at this point, so it's 20 meters per second square. That should equal g times h two. G is given, 10 meters per second square, times h two, h two is 15 meters, plus half v two squared. I don't know v two, I need to figure that out. Okay, so, let's see what we get. 20 squared is 400. 400 divided by two is 200, so let me write that directly. So I get 200, on the lefthand side, meters square per seconds square, and that equals, let's see what I get here, 10 into 15 is 150. So I get 150, again, meters square per seconds square. Oops, seconds square. Plus half v 2 squared. Okay, let's move down so that we can make some space. Now, since I want what v two is, let's subtract 150 on both sides, so now I'll get 200 minus 150 on the lefthand side, and that gives me 50. 50 meters square per seconds square on the left side. That equals half v two squared, and if I multiply by two on both sides, I will get 100 meters square per seconds square equals v two squared, and since I want what v two is, I can just take the square root on both sides, and that gives me square root of 100 is 10. 10 meters per second equals v two. There's our answer. And so, when the ball reaches the height of 15 meters, it's velocity would be 10 meters per second. It's slower, it has become slower, and that's what we would predict. All right, let's solve one more. In this problem, we'll directly look at the diagram, okay? So we have a ball that's being throw horizontally at 10 meters per second from a height of 40 meters from the ground, okay? So the ball is gonna go in some kind of a curved path. And the question now is, "With what speed will that ball go and hit the ground?" Okay so how do we calculate this velocity? At first, this problem does look a little complicated, because you see the ball is going in a curved path, so there is direction changing and everything, right? But energy conservation doesn't care about directions, okay? So even if the ball is going in a curved path, this principle works. So all we have to do is say total energy here should equal total energy over here, and we can solve it. So, great idea to pause the video and see if you can try and solve this yourself first. Okay, so again let's call this as position one, let's call this as position two, and just like before let's write down the values. So the initial steps will all be the same, let's just look at what our h and v values will be. H one would be the height at point one, that is 40, so we know that. V one is also something we know, that's 10. H two is the height at point two, which is zero, in this case. V two is the velocity at point two, hey, that's what we want to calculate, so v two is again what we want to figure out. So let's go ahead and substitute. Again, let's take g as 10 itself, so it'll be 10 meters per second square times h one this time is 40 meters plus half v one square, v one is 10 meters per second, the whole squared. That should equal, this part is zero, so I will not write that, because h two is zero, there is no height. When it's just about to hit the ground it has zero height. So zero plus half v two squared. And again, the problem reduces to algebra, we just have to simplify this, which I'm pretty sure now you can do. And if you do that, this time you will get v two as 30 meters per second. So this means the ball hits the ground with a speed of 30 meters per second. So in general whenever we are solving problems which include heights and speeds, or in general, positions and speeds, then energy conservation is a great way to solve that problem. However, the drawback of this is you cannot calculate the time. So for example if I asked how long it takes for this ball to go from here to here, you can't use this because there is no time in this equation, okay? So it's only useful as long as you don't have to worry about time.