Hey, guys. Let's work this problem out together. In this problem, you're going to throw a 6-kilogram object down from an initial height of 20 meters. Let's go ahead and draw our diagram out. Remember that's step 1 for solving conservation of energy problems. So we have our height of 0, which is our ground level, and then we have at some point here at \( y = 20 \) meters, we have a ball at 6 kilograms, and we're going to spike it down. We're not going to drop it, right? We're not going to just release it. We're actually going to spike it down with some initial speed and that's actually what we want to calculate in the problem. What's that initial speed? The only other thing we know about this problem is that when the ball finally reaches the ground right before it hits, it has a final velocity of 30 or final speed of 30 meters per second. So what ends up happening is we can choose our upward direction to be positive, and therefore, this final velocity is going to be negative. We should also expect that when we calculate the initial velocity, that should also be a negative number. Okay. So let's check this out here. We're going to use energy conservation, so we're going to have to write out our big equation for this. This is going to be \( K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}} \). So let's go through each one of our terms here. We have some kinetic energy because we have some initial speed. We also have some initial gravitational potential because we're at a height of 20. Right? So this is above our reference point \( y = 0 \) where there is no gravitational potential, so you have some stored energy here. So what about \( K_{\text{final}} \)? Well, this is going to come from the final velocity here which is we know is, is 30 meters per second or the final speed. And so we also, do we have any gravitational potential? Well, once we hit the floor, we actually have no gravitational potential because our height is 0. Alright? So let's go ahead and write out each of the terms here. I've got \( \frac{1}{2} m v_{\text{initial}}^2 + mgy_{\text{initial}} = \frac{1}{2} m v_{\text{final}}^2 \). So what we're looking for is \( v_{\text{initial}} \) and, we know that this height here \( y_{\text{initial}} \) is 20. Alright? So one thing we can do here is we can actually cancel out the masses because they appear in all the terms of the problem on the left and right side. What we can also do is, if I have fractions in some of the terms but not all of them, what I like to do is sort of multiply the equation by 2. It doesn't change anything. As long as you multiply everything by 2, nothing changes. But what you do do is sort of get rid of the fractions. So what this ends up being is you get \( v_{\text{initial}}^2 \) and then this is going to be plus \( 2g y_{\text{initial}} \) Initially, this was just \( mg y \), so you have to multiply it by 2 and it becomes \( 2g y \). So this is going to be \( v_{\text{final}}^2 \). This is \( v_{\text{final}}^2 \). Alright? So this is what we get. Now this hopefully should look a little familiar to you. This equation is really just equation number 2 sort of rewritten in a different way from back in our motion equations. So we can actually use conservation of energy to come back to the same equation that we saw when we looked at kinematics in motion. Alright? So let's go ahead and solve for this velocity here. What we're going to get is we're going to get the initial velocity squared equals we're going to move this over to the other side, and what we're going to get is, the \( v_{\text{final}}^2 - 2g y_{\text{initial}} \), and then we're just going to take the square root of this whole number here. So the answer I'm getting for \( v_{\text{initial}} \) is the square root, this is going to be \( 30^2 \). It actually doesn't matter if you plug it in as positive or negative, so you can plug this in as negative 30. The square will actually just make it and turn into a positive. Minus \( 2 \times 9.8 \times 20 \). So if you go ahead and plug this in, what you're going to get is you're actually going to get 2 answers for this. You're going to get positive or negative 22.5 meters per second. So what does that mean? It means that you could have thrown this thing either upwards or downwards. And in either case, you'll still have a final velocity of negative 30. Right? So what happens is we know that our initial velocity has to be negative because we're throwing it downwards. So, therefore, it's just going to be the negative number. So it's negative 22.5 meters per second. Alright. So that's it for this one, guys. Let's move on.