5. Projectile Motion

Positive (Upward) Launch

# Height of a Roof

Patrick Ford

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Hey, guys, let's check out this problem here. We've got a child that's throwing a ball up to a rooftop. We're gonna figure out how high is the roof. So let's go ahead. Draw diagrams. So I've got the ground level like this on then the rooftop like this. So we've got this ball that's being thrown at some angle like this, and it's gonna go up. And what else we told, we're told that the ball is gonna reach its maximum height directly above the edge of the roof. So when we go to draw diagrams, that's actually really important. We draw out the correct way. What this is saying here is that the maximum height of the trajectory is gonna be right above the edge of the roof, and then it's gonna fall back down. And so then it's gonna fall back down, and we know that this is gonna be 3 m away from the edge of the roof. So let's just go ahead and stick to the steps. We're gonna draw the X and Y pass and the points of interest and start filling out everything we know about the problem. So in the X axis we're going here and on 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 gonna look like this and in the why it's gonna look like this. All right, so that's our passing 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. All right, It's basically all we know here. Eso Let's go ahead and move on to the second step. We're gonna determine the target variable. What are we looking for It. We're looking for how high the roof is. So we look at our diagram here with that with that distance would represent is what actually represent this vertical displacement here. And if you look at it, that's basically from the ground, which where we shoot it, which we were, 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 see here. So that's actually gonna be our target Variable Delta Y A to C is gonna be the height of the roof, which actually brings us to the third step, the interval and which equations we're gonna use. Well, if we're looking for a to see, then one interval we can choose is the interval from A to C. But what happens is if you just go ahead and write all these variables, you're gonna end of figuring out that you can't actually solve this. So I'm just gonna warn you right there. If you try to do it, you're gonna get stuck. So we're gonna need a different kind of approach for solving Delta y from A to C. Now, one way you could do this. You could actually break up the motion into several parts If you wanted the Delta y from A to see, 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 gonna 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 be here. We're gonna be able to use two out of five 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 gonna use the interval from A to B instead. So I've got my A y, which is negative 9.8. We've got my V not y, which is V A y, which is Let's see, actually, let me go ahead and finish out the rest of variables. V Y is V B y. 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? All right, so let's go through each of these variables V A y. So if we have the magnitude which is 13 and the direction 67.4 so you can figure out the not X, which is V a X, which is just 13 times the co sign of 67.4. And if you do this, you're gonna get five. If you do V not why which is V A. Y? This is gonna be 13 times the sign 67.4 and you'll get 12. So we know this is 12 were here. What about V B y. The final velocity will remember. That's an important point in our project on motion problems, because we know that v. B y. Basically the velocity at the top here is equal to zero. Once you hit the top of your pique, your velocity is instantaneously zero. So it's momentarily zero for that little point there. So it's zero. So that means that we have 3 to 5 variables and we're looking for this Delta Y from A to B, which means I could just go ahead and pick the equation that ignores time. So that's what we're gonna do here. So that's equation number two, which says that the final velocity V B Y squared is V a Y squared, plus to a Y times Delta y from A to B right. We know this is equal to zero. That's why it's important and useful. We know this is 12 squared, plus two times negative, 9.8 times delta y from eight to be. If you go ahead and work this out, you're gonna get Delta Y from A to B is equal to 7.35 m. So this is only just part of the peace. 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 we're gonna dio is now that we're done with the A to be interval. We could just move onto the next one. So you can say in the B two C interval. I'm just gonna write out the same variables again. Right? So I've got negative 9.8. Now, my initial velocity from the B two c interval is actually gonna be V B Y. That's my initial velocity. We know that zero. What about B y? The final velocity That's gonna be V C Y on. 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 gonna be what we're plugging over here. And we should expect it to be negative and then t from B to C. All right, so now it looks like we have two out of five variables, but I'm gonna need either the final velocity or I'm gonna need time. So remember, whenever I get stuck in the y axis, I'm just always gonna go over to the X axis. So the easiest one to solve in the X axis is gonna be the time. So I'm just gonna go over here and I'm gonna look for time. So If I'm looking for TBC, then I'm still just gonna look at the BC Interval here. I'm just gonna write out my equation. Delta X BTC equals v X times TBC So do we have both of these other variables? Remember, we're looking for time. So do we have V X? Of course we dio because we just figured out that in the in the last part we know that's equal to five. What about Delta Extra BTC? Remember that it's 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 m from the edge off the roof, and we know that the distance from the edge to the roof over here is gonna 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, both of these variables, this is just three and then we have five times t v c, which means that T V C 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 three out of five variables 12 and three. So we're gonna go ahead and pick the one that ignores the final velocity. So we get Delta Y from B to C. I'm gonna use equation number three this time. So Delta Y from B to C is V B y. That's the initial velocity times T V C, which we know zero, because remember, this is just equal to zero plus one half times negative, 9.8 times 0.6 squared, right? So I'm just gonna go ahead and fill those in, and what you get is that Delta Y from B to C is equal to negative one point 76. So this is the vertical displacement, and it totally makes sense that we got a negative number because remember that the vertical displacement between these two points because it points downwards should be a negative number. So you get negative 1.76 here. And so when you add these two things together, you're gonna get 56 m and that is your final answers. That is the height off the building. So that's answer choice. See? Alright, guys. So we can see here. Whenever. Whenever you get stuck using one interval, you can always basically just break it up into a combination of intervals on. Go ahead and figure out what your target variable is using that method. All right, so let me know if you guys have any questions and I'll see you in the next one.

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