We've had a lot of free vectorish treatment of 2D kinematics. Now, let's do something real. Let's do the shoot and drop. This is a very well known famous demo where you have two objects released at the same time, one goes straight down, and one gets shot forward. And the age-old question is, which hits the ground or the table first? Maybe the one that drops it first. Maybe the one that got shot is moving a little faster. Let's apply some physics to it. So we have a shooting drop set up here. So it's a little spring-loaded mechanism, and one mass is hanging on a rod, and it's just going to get the rod pulled out from under, then it's going to drop. The other mass is going to be pushed by that same rod, and it's going to shoot out that way. So they'll go at the same time when I release the spring. So we can just do a quickie drawing here. So one, this will be shoot, right? So it's going to get an initial velocity like that. And one is going to be drop, and it's just going to drop like this. It's just going to drop straight down. All right, so before we get to do it, we have to do the equations. So the way I would approach any kinematics problem is write the position vector for each mass. I know you can do it without that, but it's usually good to kind of do that, right? So, rd, the position of r drop is, let's see. In the horizontal, if we pretend it's at the origin, it has motion in the horizontal, that's for the origin. Basically, we can imagine they're both of the origin are on top of each other. So the position of this one, an horizontal 0, not going to move. In the vertical, the y axis, it has an initial height, y0 or yi, I won't be old-fashioned, yi. So, yi, and then it's going to drop, it has no initial velocity, but it has an acceleration down, right? So I'll put in -1/2 gt squared, j hat. Where g is my 0.8 meters per second squared. And then shoot, let's look at rs, that's going to, let's see. In the horizontal in the x direction, it has no initial position, but it does have a velocity. S we would just say it is vi times t, and it has no acceleration. So there's it's i hat, and its horizontal j hat will also pretend they started both at the origin. And it's going to come down. Starting out at the same height, (yi- gt squared) j hat. Even if you don't need those, it's good to write them down. There will be more complicated problems where you need them. Let's see. So the question is, which hits the ground first? When do they hit? So, the y component, Tells you when they hit. I like to imagine you're talking to the equations. You ask the equations, when does it hit the ground? And it tells you, when this is 0, right? And when does this one hit the ground? When this is 0 because that's the y position. So we actually don't need the x position to answer this question. So for drop then, you'd say when this equals 0, yi, let's see. When you have yi- 1/2 gt squared = 0. And now, it's a special time. So I'm going to put a little hit down here. It'll do a star but there's a 2 in the way, so I'll just put hit. And here, when does this one hit the ground? When yi -, my God, I leave the 1/2 off there,- 1/2 gt hit squared = 0. This is a specific time now, so that's why I put a hit on it. It's when it's equal to 0. And you could solve those, and you'd see, what does this one hit the ground? T hit, for that one, no, well, it's just this goes over here, and it's 2yi, square root of 2yi over g. And you'd solve this one, then of course, it's the same thing. The answer to the age-old question is that they hit at the same time. And because in the y, they're basically doing the same thing. They're both falling. So, horizontal motion and y motion are not coupled when you're near the surface of the Earth. The horizontal has no gravitational force, no accelerations, and the y does. Therefore, they remain separate. And when you're just freely moving. So let's try it out and see what happens. So here we go. I'm going to flip the switch, and this is the one that is going to be shoot. It's going to go that way. This is the one that's going to be drop. It's going to fall straight down. And usually, the best way to tell if they hit at the same time, is with your ear. So let's listen for it here. [SOUND] I only heard one collision. I think they hit at the same time. So one quick thing we could do since this is supposed to be a real world problem here, let's do some calculations. Yi is about 0.78s. So what is this t hit? That's the square root of 2 times 0.78 m / 9.8. It should be about 0.4 seconds. So you can go back and watch the video and decide if it really took 0.4s. Whenever I do the shoot and drop demo, I do everything I can to convince you that the two masses hit at the same time. And we do all the physics to show you that the two masses hit at the same time. And I even will let students come play with it and see that the two masses hit at the same time. But no matter what I do, at least one person after class says, now if shoot were going really fast, it would have hit later, right? Inside I just scream, no, that's why I'm doing this, this explains it. But on the outside, I'd go, no, if it's going really fast, they'll still hit at the same time. And they say, what if it was an arrow? And I'd say, if it's a narrow, they'll still hit at the same time. But it was a bullet? Still hit at the same time. They will always hit at the same time. We did the demo, we did the physics. But technically, the student is right. Eventually, they won't hit at the same time, and it's actually enlightening because it happens when your model breaks down. We're always making models and we do physics. So here, what is our model? We have this region of space. We have this thing that we drop. We have this thing that we shoot. And we say gravity is always straight down. And in that case, yeah, they'll always hit at the same time. But what if you shoot it really hard? Yes, they'll still hit at the same time. What if you should it really hard? Well, eventually, the Earth is going to curve away, right? We model it as a flat Earth. It's not a flat Earth. Eventually, yeah, they won't hit at the same time. Eventually, it'll put this one into Orbit. What else might throw it off? How about air resistance? This one eventually, if you shoot it really hard, feels a lot more force due to air resistance than the one dropping. So that might slow it down in a sort of a weird way where x and y are coupled. So it's fairly complicated to think about. What about the fact that we're on a rotating Earth? There are extra forces that show up because we are not on a still Earth. We're actually spinning around. So yes, eventually, if you shoot it hard enough, they won't land at the same time. And in terms of what I just showed you, things that will happen in here and on your exam, they will land at the same time.