 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 are asked to find the speed of the ball when it has reached a height of 15 meters. So as the ball goes up, we are asked to find what's the speed, what's its speed, when it is somewhere over here at a height of 15 meters. 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, 0.1, that should equal the total energy over here. We can call this as 0.2. 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 1 and let's call this as point number 2. So we need to calculate the velocity at point number 2, right? And here's a beautiful thing about this equation. I don't care which is initial and which is final. I could have called this as 1 and this as 2. 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, which is final, okay? So let's go ahead and substitute. Potential energy is MGH because it's the potential energy due to gravity, right? So over here, potential energy at 1, we can say it is MGH1, where H-point represents the height at this point. Now we'll substitute the, you know, values later, but let's just write it down now. So this will be MGH1 plus kinetic energy at point 1 would be half MV1 square, where MV1 would be the velocity at point 1. So that should equal MGH2 plus half MV2 square, 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, H1. What is H1? H1 is the height at this point. We know that. It's 0, right? Because I'm throwing from the ground. So I know H1 is 0. I know V1 that is this number, H2, oh, I know H2, that is the height at this point, that is 15 meters, which means I need to calculate what V2 is. And that's the only thing I can, 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 H1 is 0, therefore this will be 0, plus half, half V1 square. What is V1? V1 is velocity at this point. So it's 20 meters per second square. That should equal G times H2, G is given, 10 meters per second square times H2. H2 is 15 meters, plus half V2 square, I don't know V2, I need to figure that out. Okay, so let's see what we get. 20 square is 400, 400 divided by 2 is 200, so let me write that directly. So I get 200 on the left hand side, meter square per second square. And that equals, let's see what I get here. 10 into 15 is 150. So I get 150, again, meter square per second square, oops, second square, plus half V2 square. Okay, let's move down so that we can make some space. Now since I want what we do is, let's subtract 150 on both sides. So now I'll get 200 minus 150 on the left hand side, and that gives me 50. Meter square per second square on the left side. That equals half V2 squared. And if I multiply by 2 on both sides, I will get 100 meter square per second squared equals V2 squared. And since I want what we do 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 V2. There's our answer. And so when the ball reaches the height of 15 meters, its 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 thrown horizontally at 10 meters per second from a height of 40 meters from the ground, okay? So the ball is going to 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 we are, the ball is going in a curved path. So there's 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. H1 would be the height at point one, that is 40, so we know that. V1 is also something we know, that's 10. H2 is the height at point two, which is zero in this case. V2 is the velocity at point two, that's what we want to calculate. So V2 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 H1, this time is 40 meters plus half V1 square. V1 is 10 meters per second, the whole square. That should equal, this part is zero, so I'll not write that because H2 is zero. There is no height. When it's just about to hit the ground, it has zero height. So zero plus half V2 square. 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'll get V2 as 30 meters per second. So this means the ball hits the ground with the 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.