Everyone. Welcome back. So, in this practice problem here, it's kind of complicated, but we've got a ball that's thrown from a building at 37 degrees above the horizontal. And then basically what happens is later on, it's going to hit another building that's 24 meters away at a lower height, and it's going to break the window. So let's go ahead and just draw a quick sketch of what's going on here. Basically, we've got a building, the ball that's launched up like this, but then it lands lower and breaks a window on another building that's across the way somewhere. Alright? We want to figure out how high above the ground is the window, and we're going to round it to the nearest whole number. So here's what's going on. Right? We've got the height of this building, so in other words, this distance over here. I'll call this h_b for building, and that's equal to 50 meters. What we want to do is figure out not delta y, but we want to figure out the height of the window. So h_window, which is basically the height measured from the ground. How do we actually go about doing that? Well, this is going to be a projectile motion question, and the idea here is that we have a positive launch, and we're going to try to solve this all in one interval. So let's go ahead and stick to the steps. We're going to draw our paths in the x and the y and just label our points of interest. So in the x direction, it's basically just going to go like this. It's going to go straight across like that. In the y-axis, it's going to go up, and then it's going to come back down again. And what we know is that it's going to land at a lower height, so it's going to look something like this. So here are points of interest. We've got a, we've got b, which is the peak, c, which is where it's symmetrical, and then d, which is where it hits the window. Alright? And the and the y direction is going to look like this, a, b, c, d. In the x direction, it's going to look like this, a, b, c, and then d over here. Alright? So let's get started.
So, basically, what we know here is that the height of the window, how do we actually go about solving for that? How are we going to use a projectile motion or kinematics equation to solve that? Well, we know here that the height of the window of the building that it starts from is 50. But then what happens is that as the build as the ball flies through the air, it goes up, and then it comes back down again before it hits. So, basically, what happens here is it's going to have some displacement in the y direction. That's going to be delta y. So basically, what happens is the height of this window is just going to be the height of the building that it starts from, which is the 50 plus delta y over here. Alright? But now we're going to have to look at the target variable and figure out what interval we're going to use. If you look at this, what happens is this is just delta y. This is actually delta y throughout the entire motion. This is delta y from a all the way to d. It goes back it goes up again, goes to b, comes back down through c, and then finally goes and hits over here at d. So this is really delta y from a to d. That's going to be our target variable, so delta y from a to d. So that means that we're going to choose the interval from a to d because, usually, we're going to try to figure out if we can solve all of this just using one interval instead of breaking it up. Alright? So we're going to have to label our variables in the y-axis here from a to d. So I'm going to start with delta y from a to d. That's our target variable, so we're looking for that. We need our other 4 variables. So, for example, the initial velocity, that's v_a initial. So sorry. Let's do v initial, which is actually going to be v_a, right, in the y direction. We also have v final, which is going to be v_dy. We have the acceleration in the y axis, which is always negative g, which is negative 9.8. And then we've got time. That's delta t from a to d. Now, what about that, that time? We're actually told that this projectile flies through the air, and 3 seconds later is where it hits the window. That means that our delta t is actually equal to 3. We know what that is. So, so far, we've got 2 out of these 3, out of these 5 variables. We need 3. So that means we're going to need one of these other 3 over here. Alright? This is our target variable, so we really just need one of these, the initial or the final velocity. The initial velocity is always a little bit easier to solve for because we do have some information about it. For example, we know that the angle is 37 degrees above the horizontal. So in other words, this angle over here, we know that this angle here is 37. Unfortunately, though, we actually don't know the magnitude of the initial velocity. Alright? So what we know here is that this v_ay is actually going to be v_a sine of 37. The problem is we actually don't know what this v_a is. Alright? So we're going to need one of these 2 to solve. And because we already have this equation set up, we're going to have to figure out this VA. How do we do that? Well, what happens is when we get stuck in the y axis, we're going to have to go and look at the x axis in order to solve for this. Alright? So let's go look at the x axis and see what we can solve here.
There's only one equation to look at in the x-axis, and it's basically just that delta x is equal to v_x times t. However, what happens is we're going to use the same exact interval. So in other words, the delta x from a to d is equal to v_ax times delta t from a to d. Alright. If you look at the x axis, here's what's going on. The ball flies through the air, and it basically just goes all the way across like this. And we actually know this distance over here, which actually is our delta x from a to d, it's basically the horizontal distance the ball travels, is equal to 24. We also know again that the time is just equal to 3 seconds. So, basically, what we know here is that this is 24, which equals v_ax times 3 seconds. And we can use this to solve. Basically, just divide by 3. That cancels. This is going to be divided by 3. In other words, we're going to get a is equal to v_ax. And we know from our vector decomposition that this is actually going to be v_a times the cosine of beta. Alright? In other words, it's going to be v_a times the cosine of 37. The reason I set this up is because now we have a number, and we can relate this back to the magnitude of the velocity at a. And once we can figure out this number, we can plug it back into our y-axis equations. Now we'll have our 3 out of 5 variables. That's why I have with this whole through this whole process of setting up through x. Alright. So, basically, what's going to happen here is we can divide by cosine of 37 on both sides. That cancels that out. We're going to have to divide by cosine 37 over here. Now if you go ahead and plug this into your calculators, what you're going to end up getting here is that, this is actually going to be 10. What's going to be it's a 10.01. We can we you can just round it to 10. So this is actually going to be the magnitude of the initial velocity, v_a. Now we can take this, and we can basically just plug it back into this over here. So this is going to be 10 times the sine of 37, and we know that 10 times the sine of 37 actually ends up being 6. So now we have here is we have 1, 2, 3 out of 5 variables, which means we can ignore the, the final velocity. And now we can go ahead and pick an equation to solve. A lot of steps here, and there's a lot of sort of going back and forth. But ultimately, we're just going back to whatever we do with, you know, what we would normally do with projectile motion problems. So we go over here, and the equation that ignores the final velocity is actually going to be equation number 3. So that's going to be the one that we're going to pick. Alright? So our final equation is going to be delta y from a to d is equal to the initial velocity. So this is going to be v_ay, which we know, times the time delta a from t to d, plus 1 half, we have acceleration, and then times delta t from a to d, and that's going to be squared. Alright? So a lot of variables, a lot of little sort of little subscripts and letters. Really, these are all just numbers that we know. So delta a sorry, delta y from a to d is equal to the initial velocity. We know that's just 6. That's 6 times the delta t, which is just 3, so 6 times 3, plus 1 half. This is negative 9.8, and then this is just going to be 3 squared. When you plug this in, what you're going to get is you're going to get negative 26.1 meters. Alright? So is that our final answer? Well, no. Our final answer is not delta y from a to d. This is actually just oh, sorry. Our final answer is not negative 26.1. This is actually just basically how much the ball has fallen in flight. And this actually kind of makes sense. It's fallen because we were actually told that the ball is going to hit the building or the window at a lower height than it started from. So that delta y should be negative. Alright? So now to figure out our height above the ground, I'm actually going to go ahead and pick sort of different colors for this instead of doing this in green. This is actually just going to be I'll do this, like, in purple or something. Right? So that's going to be the height that it falls because our final answer is actually going to be the height of the window. So just plug this back in. It's going to be 50 plus our delta y from a to d, which is negative 26.1. And that actually works out to 23.9. But we're going to round it, and this is going to be 24 meters. And this is our final answer. Alright? So if we round our final answer, it's going to be 24 meters, and that's answer choice c. Hopefully, that made sense. Thanks for watching, and let me know if you have any questions.