Hey, guys. We've got an interesting little example problem here. We're going to be dropping a watermelon off the Empire State Building, and then Superman is going to fly by the instance that we release it, and we're going to calculate how fast the watermelon is going once it passes Superman again on the way down. Let's check it out. So, basically, we've got a bunch of things passing each other, but now we've got this thing happening in vertical motion because we're going to have things falling. But we're still just going to follow the same exact steps that we use for catch and overtake problems.
So first things first, let's just draw the diagram and list everything we know about the problem. This is step one. We've got the Empire State Building that looks like this. Well, it's pretty good. So, we're all the way up here at the top, and we're going to drop a watermelon, and it's going to fall. But the instant that we drop it, Superman is going to be coming by, and he is essentially going to be going in the same exact direction. And eventually, the watermelon is going to catch up to Superman, so this is going to be the overtaking point.
Now, I'm going to use red for the watermelon, and my blue is going to be Superman. So now let's list what we know about each one of these objects so we can write out the position equations. That's going to be the second step. So we're dropping the watermelon from rest, which means that the initial velocity for the watermelon, which I'll call \(v_{0a}\), is equal to 0. But I know that the watermelon does accelerate, so it is actually accelerating here. But because this is vertical motion, I already know what that acceleration is going to be. I already know that the acceleration in the y-direction, so I'll call this \(\ a_{ay} \), is just equal to negative \( g \), which is going to be negative 9.8. And remember that when we do vertical motion problems, remember that the positive direction is always going to be up, and that's why it's negative 9.8. So then, eventually, this thing is going to get down here at some final position, where these two things are equal to each other, the same position at the same time. And I'm trying to figure out what the final velocity is of the watermelon. So what about Superman? Superman is going to fly down, heading at a constant 35 meters per second. So that means at this point, Superman does have an initial velocity. So \( v_{0b} \) is equal to 35. But be cautious because remember that the positive direction is upwards. So because this velocity is downwards, we actually have to put a negative sign here. You have to remember that. And so, the acceleration for Superman, so the acceleration \( a_b \) in the y-direction is actually 0 because it says gravitational force doesn't matter, doesn't apply when you're Superman. He can fly at whatever speed he wants to. We also know the last thing, that the distance of the Empire State Building is 320 meters. So what we can do is say that this is actually the 0 points, and we can say that the position that we're already starting off from, our \( y_0 \), is equal to 320, since we're actually heading downwards from it. So we can just say that \( y_0 \) is equal to 320, and it's actually going to be the same for both of them since they're starting at the same place. That's it. That's all we know about this problem.
So the second part here is we're just going to write out the full position equations for each. So the second part here is we write \( y_a \) or actually we're going to use \( y_a \) because we're talking about the vertical motion direction. So all these things still apply for vertical motion. So my \( y_a \) is going to be my initial position in the y, which is 320. That's what we said. Right? That's over here. Plus the initial velocity times time, but we know that the watermelon gets released from rest, which means that this term actually just goes away. So we've got 0 plus, and now we've got 1 half times the acceleration in the y direction, which we know is negative 9.8, times \( t^2 \). So if you simplify this by canceling out, so, you know by, doing the one half times negative 9.8, we're going to get negative 4.9 \( t^2 \) plus 320. So let's do the same thing for Superman. So \( y_b \) for Superman, 320, the same initial position plus. Now, we've got the initial velocity of negative 35, so it's negative 35 \( t \) plus, and now we've got any acceleration times time. But remember, that Superman doesn't have any acceleration. So that means this whole term goes away, so we've got 0 over here. So we rewrite out the same thing. We're going to get negative 35 \( t \) plus 320.
So that means these are the position equations over here, which bring us to the 3rd step. Now we're just going to set those position equations equal to each other. Because remember, we're trying to solve for time. That's always what we're trying to solve for in these catch and overtake problems. So I've got negative 4.9 \( t^2 \) plus 320 equals negative 35 \( t \) plus 320. So these things are actually going to have the same 320 on both sides of the equation. They're starting off in the same place. So one of the things we could already do is just cancel them both out. We're going to subtract 320 from both sides. It's basically just a wash. They basically just go both go away. And so we're only left with these last two terms, 4.9 \( t^2 \) and negative 35 \( t \). Now the other thing that we can do is we can also cancel out the negative signs that are for both of them because they're both negative. What we end up with is 4.9 \( t^2 \) equals 35 \( t \). Now remember, we're trying to solve for this \( t \) here, but notice how I have a \( t \) on both sides of the equation, which means I can cancel out one term and one term and I just end up with 4.9 \( t \) is equal to 35. And now the last thing I do is just divide by 4.9. So my \( t \) is equal to 35 over 4.9, and what I get is I get 7.1 seconds. So this is the time it takes for the watermelon to catch up back to Superman because the watermelon is accelerating. But I'm not done yet. Remember, this isn't my final answer because what I'm really looking for is I'm looking for the speed of the watermelon once it passes.
So let's take a look at our diagram again. We know now that the time it takes for both of these things to catch up to each other is 7.1 seconds. So now I just have to figure out the speed of the watermelon. So let's do just that over here. So if I want to know my \( v_a \) here, then that's a motion variable. I need to use one of my motion equations. But to do that, I need 3 out of my 5 variables. So do I have that? Well, I've got the initial velocity. I've got the acceleration in the y direction, and I also now have the time. So that means that I have 3 of my 5 variables, and I can figure out my final velocity. What this means is that my \( \Delta y_a \), basically, the distance that it takes for them to overtake each other, I don't know what that is, but it doesn't matter because I don't know and I don't care about it. So it's my ignored variable. So now that just leads us to that step in the motion equations, where we just pick any equation that doesn't have the ignored variable, which is going to be the first one. The first one relates just the final velocity, initial velocities, and the accelerations of the times. So that means that my \( v_a \) so I'm just going to pick equation 1, which says that my final velocity for \( a \) is my initial velocity for \( a \) plus my acceleration in the y direction times time. So I know that this is equal to 0, right? It starts from rest, and so my \( v_a \) is just equal to my acceleration in the y-direction which is negative 9.8. It's always it for vertical motion times time, which we now know is 7.1. So if you work this out, what you're going to get is negative 69.6 meters per second. Notice how the negative sign tells us that the watermelon is falling down and the number is higher than Superman, which actually makes sense. The watermelon has to be going faster to catch up to Superman once it passes. Anyway, guys, so this is your answer. Negative 69.6 meters per second. So notice how we can use all the same steps for vertical motion in terms of catching overtake problems. It's the same exact thing. You just go through all the steps and you get the right answer. Alright, guys. That's it for this one.