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.