AP Physics I--Romeo/Juliet Projectile Motion Problem 10-6

All right, today we're going to work on solving the Romeo and Juliet problem that we've looked at a couple of times. In class, here we are on page 66 of your textbook. Please turn there now. So we can work on it together. So you can look at it there. So we have here a ball. Sorry, wrong questions above. We have Romeo is chucking pebbles gently up to Juliet's window. And he wants the pebbles to hit the window with only a horizontal component of velocity. And he is standing at the edge of a rose garden 4.5 meters below her window, and 5.0 meters from the base of the wall. How fast are the pebbles going when they hit her window? So we need, here we have our sketch. So we have a 4.5 meter height with a 5 meter horizontal displacement. So writing in now, our given box of information, we have a Y sub F of 4.5 meters. We have an X sub F of 5 meters. In both cases, the initial position, both X and Y are going to be zero meters. And as always in these projectile motion problems, the acceleration in Y is going to be a negative G because we are going to set down to be negative on these kind of problems. And the acceleration in X is going to be zero or you can say it's negligible because it's not really zero. It's just small enough that we don't have to worry about them in our problems. Now, we know, well, we don't know the initial velocities of either. We do however know that the velocity in X is going to stay the same. So whatever the initial velocity is, is whatever velocity is going to be traveling at X when it hits that window. So that's really what we're looking for is that final X velocity. For initial X velocity. Now the initial Y velocity is unknown. We do know the final Y velocity, which I'm just going to mark as V sub Y and that is zero because as the problem states, he wants the pebbles to hit the window with only a horizontal component of velocity. So we know that is at the maximum height where the initial velocity or the Y velocity has gone to zero. And we know that time connects the two coordinate systems, the X and the Y's. So we need to figure out how fast the things are going. We know it's going to end in the initial or ends in the velocity in X so that means we're probably going to have to start with information in Y to find that time. So looking at the equations we have available to us, we know that Y sub F equals Y nought plus V nought Y T minus one half GT squared. Now we know that the initial Y position is zero. And we don't know the initial velocity and why this is what we're looking for. We do know the final position and Y and we do know the gravitational field. So plugging R numbers in that we have, we get 4.5 meters. I'm going to go ahead and leave the units off so it's a little bit cleaner to look at is equal to the unknown initial velocity in Y times time minus the one half 4.9 times T squared. So we still need to know this to be able to figure those out. So if you work your way out and you find this, don't think, oh, I failed, I don't know what I'm doing. This is a chance now for you to learn. I need to find this other information. We will come back to this. So the other equation that we've been using is assuming a board didn't get erased, is that equation right there. So we have our V sub F squared. Or V sub Y actually is what I should have written there, is equal to V nought Y squared. And because we know it's negative G for acceleration negative two, G delta Y now, we know that the velocity in Y at the top of that arc that maximum height has gone to zero. Again, that's what the problem wants to happen. So we're looking at the maximum height. So this term drops to zero. So that allows us to solve this, V not Y squared, is equal to two G delta Y, basically just added this term over to the other side, so as the negative went away, where it went. So we have V nuts Y squared is equal to two G delta Y so to get our initial velocity in Y we take the square root of two G, delta Y is going to be the square root of two times 9.81 times 4.5, and leaves the unit off there. I've been kind of in the habit there on these, so they're a little bit cleaner to see. So we have V nought Y plug that into the calculator. So you're going to get the square root of two times 9.81 times 4.5, hit execute. We have 9.39 rounding that to the is it two significant figures that we're working with on this problem? Let's take it back a quick look at what we're given. 4.5, 5.0. So yes, we're only dealing with two significant figures. So 9.39 becomes 9.4 meters per second, and this is in the Y direction. The initial Y velocity. As an answer, a question, because this is how fast Romeo threw the balls in the Y direction as pebbles. We don't yet know how fast you threw them in the X direction or how fast they hit the window. So we've only done part of the problem. We now need to use this information to find the time, so coming back up here, we can now substitute this in as 4.5 equals 9.4 T minus and one half of 4.9 or actually that should have been 9.8 up there. Sorry about that guys. I took already divided by the two. So 9.8 divided by two there should get us 4.9. T squared. And now we have our quadratic. So we can plug that into our calculator. I'm going to put it into the standard form first. So we're going to get 4.9 T squared. -9.4 T plus 4.5 equals zero. I just added to the 4.9 T squared to the other side, then subtract the 9.4 over here so that we have the standard form. On your calculators. To solve a quadratic unless you are just really want to do the quadratic formula by hand today. You can do that. Press math, press the up arrow and you'll be taken to solver, select that. Make sure you're on a screen that shows equation solver at the top. Enter in the equation. Let's clear out what we have here. And we're going to have 4.9 T squared. Variable is kind of default to X on most calculators. -9.4 T plus 4.5. Double check to make sure I've entered that in 4.9 T squared -9.4 plus 4.5. I hit enter. This is not our answer. This is random from whatever I was doing on this calculator. The last time I used the solver hit alpha solve, and we get .918367 three four 6 9 three, Ken were running to two sig figs, so we're going to get .92. So we have a time of .92. For those of you using calcios and you use the equation solver on there, you have those two answers. And you know the .92 is the correct one because the other answer is most likely a negative number. And we can't have a negative value on these. Let me double check. Okay. So now we have a time. In so once we get to time, we can take our information from the Y, we can take that time, plug it into X equations, and we can figure out what we're actually trying to solve in this problem. So our equation is again that one right there, assuming the board didn't get a race while I was gone. We have X sub F is equal to X zero plus V nought XT. Plus one half a sub X T squared. We know that X nought our initial X position was zero. We know that our acceleration in the X direction is zero. So we're left with X sub F equals T we have the time. We have the X position. So solving for V nought X or just V sub X as it is now. Since it doesn't change, we get X sub F over T, which X sub F is 5.0 meters. Our time we solve for earlier was .92 seconds. So taking your calculator making sure that you leave solver mode first by hitting mode or second quit. We now take our 5.0. Divide it by .92, and we should get 5.4. Meters per second. And that was the velocity and how fast it was going as it hit the window.