Skip to main content

ï»¿ >> At some point, I'm going to get like the little applause light, you know, that comes on, and like laugh, you know. Hello class, Professor Anderson here. Welcome to another installment of the looking-glass physics lectures. I'd like to talk to about two-dimensional motion. We've learned a lot so far about one-dimensional motion and how the kinematic equations can tell us what the velocity is if you're given an acceleration, or what the position is on the function of time if you're given the acceleration. I would like to extend this idea to two dimensions. We, of course, live in a three-dimensional world, but a lot of the problems that you're going to face in physics are really two-dimensional problems. So for instance, if you throw an object, that object actually follows two-dimensional motion. It has some X position, it has some Y position, but it doesn't really come towards you or away from you so there's no third-dimension there. It's really a 2-D problem. All right, so let's talk about two-dimensional kinematics. And let's approach this with the following thought experiment. Let's say that we go to the top of the Empire State building and we take an object, say a marble, and we throw it off the top of the Empire State building, and we throw it off horizontally. You, of course, should never do this. I'm sure you'll get in big trouble if you do. But if I launch this object horizontally, what sort of trajectory does it take? Anybody know what that trajectory is called? Yeah? >> A parabola? >> A parabola, right? It sort of curves down, like so. And that is, of course, a parabola. Is there another name for this projectile after it leaves my hand? Is there another name for that type of motion? And I'll give you a hint. I just gave you the two words that are in the answer. Yeah. >> Projectile motion? >> Projectile motion, okay? Just this is projectile motion. We have launched an object. It follows a parabola, and it is projectile motion. That's what we call it. All right, so let's see if we can understand a little bit about this motion. This is, of course, the motion that governs a lot of things as they move around near the surface of the earth. And so it would be nice to understand the problem. So first off, let's draw a coordinate system. Let's say that Y equals zero is down here at the ground. You can, of course, make that wherever you want, but that seems like a natural starting point. Where we launch the object from, let's say that is Y equals H. All right, now what do we do? Well, let's think about this motion for a second. First off, why is it curving down? What is dragging this thing down towards the earth? Yeah? >> Gravity? >> Gravity. Gravity is, of course, pulling it down towards the earth. No gravity -- this thing just keeps going on forever and ever. And so gravity is really important here, and in fact, the acceleration due to gravity we know. It is negative G, which is negative 9.8 meters per second squared. What about the other dimension? What about the X dimension? What is the acceleration in that direction? Does anybody know? Yeah, over here on the left? >> Zero. >> Zero. Why is it zero? >> Because X is going flat? >> Because X is a horizontal, okay? And as we're going to learn later, there is no force in the X direction. The only force here is gravity, which is downwards. And so in fact, there is no acceleration at all in the X direction if we're ignoring air resistance, which we are. Okay, this set of conditions is projectile motion, or also known as freefall motion. When you are only under the influence of gravity, then you are in freefall. So now what do we do? Well, let's ask the question, how long is this thing in the air? How long is that thing in the air? How do we deal with this problem? Well, we've learned a lot about motion in one-dimension. We've learned about the kinematic equations. Let's just go back to the kinematic equations for a second and see what we have. All right, what we have is equations like this: X final equals X initial plus VX initial times T plus 1/2 A sub X T squared. All right, that looks pretty good. But X is just some variable, right? It could be whatever variable I want. So let's just change it to something else. Let's change all of those Xs to Ys. Y final equals Y initial plus VY initial times T plus 1/2 A sub Y T squared. That seems perfectly valid. Okay and now, I like this because we usually say the vertical position is Y. And in fact, we said Y equals zero is down there and Y equals H is up there. So this equation looks like it might be able to help us answer the question, how long is this thing in the air? All right, so let's take a look at this equation. In our problem, what is Y final? Yeah, what's Y final? >> It's the ground? >> It's the ground. And we call the ground zero, all right. We could have called something else zero, but the ground seems like a perfectly good choice for the height of zero. If the ground is zero, then where we started initially is, of course, H. What about VY initial? And let's be clear here, that we are going to launch this object horizontally, okay? That's kind of key in our discussion at this point. If we have launched this thing horizontally, what is the vertical component of the velocity? Yeah? >> That would be zero as well. >> Be zero, right? It's sort of actually pretty hard to launch something perfectly horizontally, but let's say we do launch it perfectly horizontally. Then that term is zero as well. And now, we're just left with this, 1/2 AT squared. We said A was minus G, so I'm going to pull that out in front, and I get minus 1/2 GT squared. And now, look. I can solve this equation for T. I'm going to move this over to the other side. I get 1/2 GT squared equals H. And now, I can solve for T. I multiply across by 2. I divide by G. And I take the square root. Okay? This is the time it takes for any object to fall from a height H. And there's something sort of interesting here, which is the following. Let's say we didn't just launch it horizontally. We just stood there and dropped it vertically. Would this time be exactly the same? Yes, because this equation right here is only dependent on the height that you started from. Since we have zero vertical component of the initial velocity, we would get the exact same time. And now, we can plug in some numbers and try it out. Okay, so how high is the Empire State building? Somebody told me this earlier. They looked it up on their phone. How high was the Empire State building? >> It's about 1250 feet. >> 1250 feet, okay. And that is, of course, English units. We like SI units, so let's just approximate that as 400 meters. You can do your calculation yourself and see if that's pretty close, but I think that's pretty reasonable. And now let's see if we can calculate the time. T we said was square root of 2H over G. So it's 2 times 400 meters divided by G, 9.8 meters per second squared. And now, we can run this in our calculator or we can just approximate it. 2 times 400 is 800. 800 over 9.8 is pretty close to 800 over 10, which would be 80. Square root of 80 is about 9. So I would say 9. Did anybody run it on their calculator and get an exact answer? Yeah? What did you get? >> 9.035. >> 9.035 seconds, okay. So pretty good guess we had for 9 seconds. And that's how long that thing would be in the air. Okay, 9 seconds. All right, hopefully that one is reasonably clear. If not, definitely come see me in office hours. Cheers.

Related Videos

Related Practice

© 1996–2023 Pearson All rights reserved.