Hey, guys. Let's check out this problem here. We've got a child who is throwing a ball up to a rooftop and we're going to figure out how high the roof is. So let's go ahead and draw a diagram. I've got the ground level like this, and then the rooftop like this. We've got this ball that's being thrown at some angle like this, and it's going to go up, and what else are we told? We're told that the ball is going to reach its maximum height directly above the edge of the roof. So when we go to draw our diagrams, it's actually really important that we draw it the correct way. This is saying here that the maximum height of the trajectory is going to be right above the edge of the roof, and then it's going to fall back down, and it's going to fall back down, and we know that this is going to be 3 meters away from the edge of the roof. So let's just go ahead and stick to the steps. We're going to draw the x and y paths and the points of interest and then start filling out everything we know about the problem. So in the x-axis, we're going here, and in the y-axis, we're going up to the maximum height and then back down to this point right here. So our points of interest are a, the initial, b, which is the maximum height, and then finally when it hits the rooftop which is at point c. So in our x, it's going to look like this, and in the y, it's going to look like this. Alright. So that's our path in the x and y. And let's see. What else do we know? We know the initial velocity here is 13. We know the angle is 67.4 degrees. Alright. It's basically all we know here. So let's go ahead and move on to the second step. We're going to determine the target variable. What are we looking for? We're looking for how high the roof is. So if we look at our diagram here, what that distance would represent is it would actually represent this vertical displacement here. And if you look at it, that's basically from the ground where the child launches the ball from, and then to the place where it hits the roof. So in the y-axis, what we're really looking for is we're looking for the distance between a and c. So this is delta y from a to c here. So that's actually going to be our target variable. Delta y from a to c is going to be the height of the roof, which actually brings us to the third step, the interval and which equations we're going to use. Well, if we're looking for a to c, then one interval we can choose is the interval from a to c, but what happens is if you just go ahead and write out all these variables, you're going to end up figuring out that you can't actually solve this. So I'm just going to warn you right there, if you try to do it, you're going to get stuck. So we're going to need a different kind of approach for solving delta y from a to c. Now one way you could do this is you can actually break up the motion into several parts. If you wanted the delta y from a to c, that would actually just be the distance from a to b, right? So this would be delta y from a to b plus the downward displacement from b to c. You could basically, if you could figure out both of these vertical displacements, then you would be able to add them together, and that would represent the difference between them, delta y from a to c. So this is actually going to be the approach that we could take for this problem. Delta y from a to c is delta y from a to b plus delta y from b to c.
The reason this is useful is because we know a lot of information about this point b here. We're going to be able to use 2 out of 5 variables that we already know. So basically, I just have to figure out both of these vertical displacements, add them together, and then that will be my final answer. So that brings us down to the next step. So we can't use the interval from a to c, but if we want to figure out delta y from a to b, then I'm just going to use the interval from a to b instead. So I've got my Ay which is -9.8. We've got my v0y, which is vay, which is, let's see. Actually, let me go ahead and finish out the rest of the variables. Vy is vby, and then we've got delta y from a to b, and then we've got t from a to b.
So what's the time? Alright. So let's go through each of these variables, vay. So if we have the magnitude which is 13 and the direction 67.4, so we can figure out v0x which is vax which is just 13 times the cosine of 67.4. And if you do this, you're going to get 5. If you do v0y which is vay, this is going to be 13 times the sine 67.4 and you'll get 12. So we know this is 12 over here. What about vby, the final velocity? Well, remember that's an important point in our projectile motion problems because we know that vby, basically your velocity at the top here is equal to 0. Once you hit the top of your peak, your velocity is instantaneously 0. So it's momentarily 0 for that little point there, so that's 0. So that means that we have 3 out of 5 variables, and we're looking for this Delta Y from A to B, which means I can just go ahead and pick the equation that ignores time. So that's what we're going to do here. So that's equation number 2, which says that the final velocity vby squared is vay squared plus 2ay times delta y from a to b, right? We know this is equal to 0. That's why it's important and useful. We know this is 12 squared plus 2 times -9.8 times Delta y from a to b. If you go ahead and work this out, you're going to get Delta y from a to b is equal to 7.35 meters. So this is only just part of the piece. Remember this distance here is just only one of the pieces of our equation, so we need 7.35 over here, and then we just need to figure out delta y from b to c. Right?
So what we're going to do is now that we're done with the A to B interval, we can just move on to the next one. So we can say in the B to C interval, I'm just going to write out the same variables again. Right? So I've got -9.8. Now my initial velocity from the B to C interval is going to be vby. That's my initial velocity. We know that's 0. What about vy, the final velocity? That's going to be vcy, and we don't know what that is, and then delta y from b to c, which is what we're trying to find. Remember, that's going to be what we plug in over here, and we should expect it to be negative, and then t from b to c. Alright. So now it looks like we have 2 out of 5 variables, but I'm going to need either the final velocity or I'm going to need time. So remember, whenever I get stuck in the y-axis, I'm just always going to go over to the x-axis. So the easiest one to solve in the x-axis is going to be the time. So I'm just going to go over here, and I'm going to look for TBC, then I'm still just going to look at the BC interval here. I'm just going to write out my equation, Delta XBC equals Vx times TBC. So do we have both of these other variables? Remember, we're looking for time, so do we have vx? Of course, we do. So because we just figured that out in the last part, we know that's equal to 5. What about delta x from b to c? Remember that was the horizontal displacement from b to c, that would actually represent this distance right here from b to c. We actually know what that is because remember that we're told the ball lands 3 meters from the edge of the roof, and we know that the distance from the edge to the roof over here is going to be the horizontal displacement from b to c. So that's delta x. So this is actually what delta x from b to c is. You should write that down. So let's move on. So now we actually have both of these variables. This is just 3, and then we have 5 times TBC, which means that TBC is equal to 0.6. So now that we have this 0.6, then we can go back to our equations in the y-axis, and we have 3 out of 5 variables, 123. So we're going to go ahead and pick the one that ignores the final velocity, so we get delta y from b to c. I'm going to use equation number 3 this time. So delta y from b to c is vy as the initial velocity times tvc which we know is 0 because remember this is just equal to 0, plus 1 half times -9.8 times 0.6 squared. Right? So I'm just going to go ahead and fill those in, and what you get is that delta y from b to c is equal to -1.76. So this is the vertical displacement, and it totally makes sense that we got a negative number because remember, the vertical displacement between these two points because it points downwards should be a negative number. So we get -1.76 here. So when you add these two things together, you're going to get 5.6 meters. And that is your final answers. That is the height of the building. So that's answer choice c.
Alright guys. So we can see here whenever you get stuck using one interval, you can always basically just break it up into a combination of intervals and then go ahead and figure out what your target variable is using that method. Alright? So let me know if you have any questions, and I'll see you in the next one.