SPEAKER 1: Couple more questions we can answer about shoot and drop. How much further did the shoot go than the drop? And at what angle did shoot hit the table? So let's see. This is one of those two body problems, although this one is easier, because one of them didn't move in x, right? So how much further along the x-axis? So drop. The dropped mass is just zero. So the position vector is pretty simple for that one. It did not move in x. And shoot. We think back to the x component of that position vector. Basically, it had zero initial position, and it just moved with constant velocity. So it would be at v. I think we called it v initial, the initial horizontal velocity, times t hit, because this is how far it got before it hit the table. So it's how much further shoot was than drop at the time that they hit the table. So we could then say the difference of those two, of course, is just this value, because that one is zero. And you could plug in things that you know. The initial times the square root of 2y initial over g. But you can't solve for it, because we don't know how fast it was going. So what are we going to do? Well, that's why the ruler's here. I saw it hit at 1.3 meters. All right? So instead of calculating how much further it went, we just measured how much further it went. For fun, let's calculate the initial velocity. So let's see. If this is 1.3 meters vi, and this was 0.4 seconds-- we did that last time. So 0.4 seconds. So in the end, you get that the initial velocity was vi, was 3.25 meters per second. Not exactly what it asked, right? It asked about this number, 1.3. But we were able to calculate, for fun, 3.5. So that was a fairly straightforward application of kinematics. Here's a weird one. At what angle? So that is a little unusual. Let's draw shoot's trajectory here for a second, and see if we can figure out what they mean. So it's a little cube, and it shot like that, and it kind of hit the ground like that. And the question is, at what angle, theta, does it hit the ground? So to get that, you've got to think about what's really defining this trajectory. As you go from point to point, it's those little delta r vectors. Right? As you go from here to here, that's a delta r. At the beginning, delta r was flat. It was moving that way. But the quantity that gets you that delta r is actually v. Remember that velocity is delta r over delta t. Average velocity. Or instantaneous is the same thing, just with a really small delta r. So it's really the velocity that tells you the direction of the trajectory. So if you needed an angle for the trajectory, you got to look to the velocity. Now how you get it is you break the velocity into components at this time, right? So really, the velocity at this time is along the trajectory, and it's down like that. It's the velocity of shoot. And it has a component down like that, and it has a component horizontal like that. And the theta you want is here. So you do your opposite side angle, whatever it is from junior high. I don't remember. Basically, this theta equals that theta. That's what you care about. So there is your theta. And then you apply tangent, right? Tangent is opposite over hypotenuse. You remember that tangent, in this case, the way I set it up, is vy over vx. The tangent of theta is the ratio of the two velocity components. Depending on where this angle you're looking for is, you might be shooting it weird, you might sometimes get vx over vy. Usually, though, it'll come out vy over vx. So let's see then if we could actually get that angle. Let's see. So the angle, just as a summary, is of the trajectory from delta r, which is the same direction as v. So we're getting it from v. All right. So let's see. So tangent theta. And let's see. What numbers do we have here? vy over vx. vx, we just figured out, because we had the initial horizontal velocity, and it doesn't change. There's no acceleration in the x. So we put 3.25 in the bottom, right? But now we need vy. Oh, let's see. It accelerated over a distance. So we've got to go back to that equation, v final squared equals v initial squared, which was zero, plus 2ad, plus 2 times a, 9.8, times d, 0.78 meters. That's how high it was, remember? Let's see. So you worked all that out. You multiply that, take the square root, and you get 3.9 meters per second. So vy, when it hits-- 3.9 meters per second. So 3.9 here. You get something a little bit bigger than 1. Inverse tangent of whatever that is, a little bigger than 1, and you get theta is 50 degrees. As you'd expect, it landed when the two components were about the same, so it's near 45 degrees. It's 50 degrees. So if you watch the video, maybe you can tell it really was at 50. So this is how you can use kinematics to get much more than just where something ended up.