Hey, guys. Let's work out this problem together. So we've got this catapult that's launching a stone. It's going to go up and then it's going to hit a castle wall. We're trying to figure out the direction of the stone's velocity. So let's just go ahead and stick to the steps. We're going to draw the paths in the x and y and then label our points of interest. When you go to do this step, you might actually realize that there are two different ways to draw the path of the stone. One possibility is that you could hit the castle wall while still on the way upwards in which the final velocity is going to be upwards like this. Or the other situation is that the catapult might be launched upwards, reach its maximum height, and then actually hit the castle wall on the way back down. And we don't know from the situation or from the information that's given in the problem which of these situations is the actual correct one. But it turns out that it doesn't really matter for our purposes. So we're going to draw the path in the x and y. So the x path is going to look the same for both of the, it's just that one of them is going to be a little bit shorter, but the y path might look different. The y path for this one is just going to go straight up, and for this one, it's going to go up and then come back down again. So when we label our points of interest, there are actually two possibilities. You could just be going from a to b, in which case your interval is going to look like this, a to b, or you could be going from a to b and then back down to c in which you hit the castle wall. So you're going to be looking like this, a to b to c. So because we actually don't really know, you should always just try to draw the passing the x and y. And if you can't tell, you can't actually figure out which one it is, it doesn't really matter because we can actually just, all we're looking for here is we're just looking for the point which it's launched to the point where it hits the wall. So I'm going to call this interval even though I don't know which one a to c. So what we're really looking for here is we're looking for the angle that this velocity makes with the horizontal. And so remember, there are two components. We've got vcx and vcy. And really, what happens is because we know the x velocity is never going to change, what's really going to affect this is the y velocity. So let's go ahead and go to the second step here, which is our target variable. What are we looking for? We're looking for the direction which is θ, and we're looking for the direction at point c. So that's going to be this variable over here. Alright. So because θc is actually just a vector equation, it's just this, tangent inverse over here, I'm going to start with that. So we have tangent inverse of and we need vcy over vcx. So vcx is the easy one because remember that the x component of the velocity never changes. So what I can do here is say that my vcx is just going to be vx throughout the whole entire motion. And that's actually just the initial velocity times the cosine of θ, right, which I know that's just 50 times the cosine of 56, and I get 28. So that's never going to change. It's always going to be 28 throughout the entire motion there. Vcy is where it gets a little tricky because I don't know what this vcy is, but I'm going to have to go find it. I'm going to have to pick an interval and then solve it using some UAM equations. So that's the third step. So our interval that we said we're going to use is just the interval from a to c. So even though we don't know what the actual trajectory is, we don't know if it goes up or comes up and back down again, it doesn't really matter because all we're really looking for is the point where it's launched and the point where it hits the castle wall. So let's just go ahead and list out all of our variables here. In the y-axis I've got ay is -9.8. I've got v0y, which is vay, which I can actually figure out here. My vay is just going to be, or rather my vinitial y is going to be vay, and that's v0 times the sine θ. So that's just 50 times the sine of 56, and that is 41.5. Man. So 41.5. Okay. So we know this is 41.5. And then our vfinal y is going to be vcy. This is actually what we're trying to look for. And then see, we've got Δy from a to c and then t from a to c. We do know that it takes 6 seconds for the stone to hit the castle wall, so that's tac. So that means that we have three to five variables, and we pick the equation that ignores my Δy. So this equation here is actually going to be equation number 1, which says that the final velocity, vcy, is vay + ay times tac. So vcy equals our initial velocity in the y-axis, 41.5 plus -9.8 times how long it was in the air, 6 seconds. If you go ahead and work this out, you're going to get -17.3 meters per second. So this is our y components of our velocity. Now remember, this is not our final answer because what we need to do is we need to take this number and we need to plug it back into this equation over here, to get our tangent inverse. So what we're going to do here is θc equals the tangent inverse of we've got -17.3 divided by 28. That's our x components. And if you go ahead and work this out, what you're going to get is you're going to get -31.7 degrees. So which one of our two situations did it end up being? Well, because the y component ends up being negative, it's actually going to be this situation over here. This was the correct one. And so this angle over here, θc, because it's negative, means it's being below the horizontal. So your θc is -31.7 degrees, and that is your final answer. Alright. So that means that brings us to answer choice A. And let me know if you guys have any questions about that one.
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5. Projectile Motion
Positive (Upward) Launch
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