Everyone, welcome back. So by now, we've seen how to solve projectile motion problems. And in some cases, instead of launching an object upwards or even horizontally, you may actually launch objects downwards. For example, if you threw a rock downwards from a building. Now don't worry because you're going to solve these problems the exact same way. It's just another variation of projectile motion. We'll use the same exact steps and equations to solve. The only thing that's different about these problems is that when you launch objects downwards, when you separate their initial velocity into the x and y axes, that \( v_0y \) will actually always be negative. It will always be. And it's basically just because of the way that we choose the upward direction to be positive. So let me just go ahead and walk you through this example. We'll see that it's really not so different from how we solve any other projectile motion problem. Let's get started.

So in this problem here, we have a rock that's thrown at 5 meters per second at an angle of 37 degrees downward. In other words, the \( v_0 \) here is 5, and the angle from the horizontal is 37 degrees downward. So because this angle here is actually pointing down and this angle is actually going to be clockwise from horizontal, this angle is actually going to be negative 37 degrees. That's important there. We're told that it hits 10 meters from the building, and we're going to calculate the height of the building. Let's go ahead and just stick to the steps here. First thing we're going to do is draw the path in the x and y and label the points of interest. In other words, this ball really in the horizontal is going to go all the way over here. The path is going to look something like this where it's going to sort of curve downward, and that means that the path in the y-axis really just goes from the building all the way down to the ground. So these points of interest in the past is actually pretty straightforward. So where the point of interest is really just the initial, we label that as A. Now there's no other sort of max height. It doesn't curve upwards or downwards or something like this. It only really just goes from the initial, and then it just hits the ground. So there's really only two points of interest in this problem, which makes setting up our interval and our pass pretty straightforward.

Okay, so now let's figure out the target variable. We're told a couple of other pieces of information here. One of them is that this thing lands at 10 meters from the building, which is really just a \( \Delta x \). Right? So \( \Delta x \) is 10. But that's not what we want. We want the height of the building, and that's really just this vertical displacement over here. So this is basically the full height the rock falls as you're going from the roof up down to the ground. So this is going to be my height, which is really just \( \Delta y \) in this problem. And so that's really what we're looking for. It's actually really going to be the absolute value of \( \Delta y \) because the height is always just a positive number.

Okay, so that's the target variable. What about our interval and our UAM equation? Well, if we're looking for \( \Delta y \) in this problem then we're actually just going to take a look at our interval from A all the way to B. That's from the rooftop all the way down to the ground. So we're really just trying to figure out \( \Delta y \) from A to B because we're actually just looking at the interval from A to B. So this is going to be A to B. That's my interval over here. So what are my variables? Well, I've got \( \Delta y \) from A to B. That's what I really what I'm trying to solve for. Well, let's list out the other variables. I need the initial velocity in the y direction. I don't have that but I can find it because I have the initial velocity and the angle. What about the final velocity? I don't know what the velocity is in the y direction when it hits the ground. What else? So I've got so let's see. I've got my \( a \) in the y direction, which is equal just to negative \( g \), which is just negative 9.8. And then what about the time, the time that it takes in the air? That's \( \Delta t \) from A to B. I actually don't know anything about the time either. So I've got some work to do here because I have to figure out 3 out of 5 variables. Alright? Let's start off with the initial velocity in the y-axis because we can actually figure that out pretty straightforward. So if I want to split this out and get the x and y velocities over here, that's really just going to be \( v_0 \) . It's actually just going to be \( v_{a\_y} \) sorry. \( v_{ax} \), which is really just going to be \( v_a \times \cos \) of this angle. So the cosine of the angle here is going to be 37 degrees. In other words, that's just going to be 5 times the cosine of negative 37. If you go ahead and plug this in, what you're actually going to get is you're actually going to get 4. And then if you do the same thing for the y-axis, your \( v_0y \), this is really just going to be \( v_a \cos(\theta) \). In other words, this is really just going to be 5 I'm sorry. This is sine. My bad. This is going to be the sine of that angle. This is really just going to be 5 times the sine of negative 37 degrees. And when you plug this into your calculator, what you're actually going to get is an initial velocity that is negative 3. That's exactly what I meant over here by your initial velocity always being negative when you actually split it up into its components. And what's really cool about this is that you never have to, like, remember to stick a negative sign. When you plug in magnitude and direction, this will always actually just give you the appropriate sign. So it's negative 3.

Alright, so this is going to be negative 3. You still don't know anything about the initial about the final velocity in the y-axis. So I really just need one other variable. It's either going to be the final velocity or the time. But one of the things I want you to remember here is that whenever you sort of get stuck in the x and y axis, you should always just go to the other axis to solve. So if you're using if you get stuck in an x-axis, solve it with a y-axis equation and then vice versa. So we got stuck here with the y-axis. This is our y-axis equations. So what I'm going to do here is to figure out the time, I'm going to go ahead and look at the x-axis. So if I go to the x-axis over here, there's really only 3 variables because I just have \( \Delta x = v_x t \). Okay? So in other words, I've got \( \Delta x = v_x t \). And so, really, what happens is I know what \( \Delta x \) is along this interval. This is \( \Delta x \) from A to B. This is really just going to be the initial velocity, which is really just the steady velocity throughout the whole problem because there's no acceleration in the x-axis. So in other words, it's just \( b_x, v_{ax} \times \Delta t \) from A to B. Okay? So we actually know what this \( \Delta x \) is. We also know what the initial velocity is in the x-axis, so we can really easily solve for this \( t_{ab} \). Okay? So, really, what this is going to be is just going to be \( \Delta x \) from A to B divided by \( v_x \). That's going to equal \( t_{ab} \). And that's really just going to be the 10 meters that it travels in the x-axis divided by its velocity in the x-axis, which we just figured out over here was just 4. 4 meters per second positive. Right? It's along the x-axis. So this is really just going to be 2.5 seconds. And that means that now that we figured out what this \( t_{ab} \) is, you can bring it back into this equation over here because we know this is 2.5. Okay? So what is now all we have to do is really just figure out which equation do we use to solve, and it's actually just going to be the third one. It's going to be the one that ignores the final velocity, which is always going to be that third equation. Okay? So what I'm going to do here is now that I'm done, I'm going to pick equation number 3. So \( \Delta y \) from A to B, which is exactly what I'm looking for over here, is just going to be the initial velocity, \( v_0yt_{ab} \), plus one half. This is going to be \( a_y \), and this is going to be \( t_{ab}^2 \). So, in other words, this is really just going to be my initial velocity, which is negative 3. So this is negative 3 times the \( t_{ab} \), which is 2.5 plus 1 half. This is going to be negative 9.8, and then this is going to be 2.5 squared again. When you plug all of this in, just be really careful. Remember, you have negative signs here and here. What you're actually going to see here is that this is equal to negative 38.1 meters. So in other words, the displacement of the rock from the rooftop down to the ground is negative 38.1 meters.

Now, what we said is you could probably just leave your answer like this. But if your professors sort of like a stickler about signs and just wants you to pay attention to the signs very carefully, what you could say is that the height of the building, which is really just the absolute value of \( \Delta y \) from A to B, is actually just 38.1 meters. So either way you represent this answer, this is probably the more technically correct one, but that's just sort of a small point there. Alright? So that's really it for this one. Let me know if you have any questions, and let's get some practice.