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Anderson Video - The Forces from Jumping

Professor Anderson
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>> Hello, class. Professor Anderson here. Let's take a look at a problem which is a person jumping up in the air. So, the way a person jumps up in the air, of course, is they push up with their legs, at the end of that extension they float through the air for a little while, they come back down and they recompress their legs. So, let's model that jump with a force, as a function of time, that looks like this. The person is just standing there and they have some force on the ground. When they jump they exert a positive force. While they are in the air the force on them is 0 and while they are coming down they land and they recompress the force plate that they're standing on and it's perhaps a different value than the initial one. Okay? So, let's give this some real numbers. When they are just standing on this force plate. Let's say their force is 500 Newtons, okay? So, that is their weight. So, that's the first thing we can understand. The weight is, of course, just equal to MG. Let's call that weight W. So, if we know the weight, W, we can solve for the mass. Mass is just their weight over gravity. Now, when they are jumping up and they're starting to push off with their legs, that force goes up and let's say it goes up to 2000 Newtons. And now at the end of that they're in the air, so there is no force on the force plate that they were standing on. So, the force is 0 that whole time. When they come back down and land they start to recompress the force plate and let's give that a number of maybe 3500 Newtons. Now, let's give a couple more givens here. This is the time while they are jumping up. This is the time while they are landing. So, this is when their legs are pushing them up, this middle section is when they are in mid-air and that last part is when they are compressing their legs as they land. Okay? Now, let's ask some questions about this process. Okay. Let's ask the following question, "What is their acceleration going up?" Okay. While they are pushing themselves off the ground, what is that vertical acceleration? We know it has to be a positive number, because they're going to jump up into the air. But how do we figure out what that is? Well, let's go to our free body diagram, okay? If we think about the free body diagram, then when they are pushing themselves up off the ground there is a force of jump, which is equal to, in this case, 2000 Newtons. But there's also a force down, which is equal to their weight. Weight is, of course, equal to MG and we know what that weight is from the earlier numbers, 500 Newtons. All right, so now we can write sum of the forces in the vertical direction is what? F jump minus MG. And all of that has to be equal to the mass times the acceleration going up. All right? So, now we can solve that equation for the acceleration going up. And if we do that, what do we get? We get A up equals F jump minus MG, all over M. And now we know all those numbers. F jump, we are given that from the graph. MG is, of course, just their weight, W. M is apparently their weight divided by G. And now we can plug in all those numbers. F jump was 2000 Newtons. Weight was 500 Newtons. We're going to divide by the weight 500 over G, which is 9.8. We're in SI units here. So, we end up with a number that looks like 29.4 meters per second squared. Okay, so that's their acceleration going up. Let's ask the question now, "What is their acceleration while they are in mid-air?" Well, you probably know the answer to this already, but let's be clear. While they are in mid-air there is only one force that's acting on them and that's gravity. And that force is in the negative direction. That equals the mass times their acceleration while they're in mid-air. And, so, A air, like you suspect, is just negative G. Negative 9.8 meters per second squared. What about when they are landing? Well, we can use almost the exact same equation, because if I think about the free body diagram while they're landing, the force of those legs is still up, gravity is still down and, so, that difference is going to help you decelerate and come back to a stop. So, in fact, we can use the exact same equation. A land is equal to the force during landing minus weight, all over W over G. And we have all those numbers from the graph. F land was 3500 Newtons, so you can plug in those numbers and if you plug in those numbers you should get 58.8 meters per second squared. All right. Let's ask two final questions on this problem. Let's say while they are jumping up they start from rest, they're standing on the ground, and we want to calculate the final speed as they just leave the ground, okay, or as they just leave that force plate. So if they're just leaving the force plate, what is their speed? All right. For that we can go back to our kinematic equations. Okay. And we know that vf equals vi plus at. vi is 0. She starts from rest on the ground. And so vf in this case is just the acceleration going up times the time that it took her to do that jump. And we have those numbers. Right. The acceleration going up we said was 29.4 meters per second squared. And how long did it take her to do that jump? Well, we're going to say it's 0.25 seconds. And if you run those numbers, you should get 7.35 meters per second. Let's ask one final question. How high does that person jump? How high is the jump? All right. Well, again, we can go back to the kinematic equations, and if we think of the person launching themselves, they're going to go up to a height h if they left the ground at v, and we know what that v is. We have it right here. Okay. Let's actually just call it v. We won't call it v sub f because we don't want to be confused when we write this equation over here. vf squared equals vi squared plus 2 times the acceleration in the y direction, yf minus yi. That's one of our kinematic equations. vf squared is 0. When you get to the top of that jump, you're moving at a speed of 0. vi squared is just this v that we just solved for, so we'll call that v. And then we have acceleration due to gravity, which is negative g, and we have the height h, right. y initial is 0, so y final is just h. And now look, you can solve this very quickly for h. We get h is equal to v squared over 2g, and now you know all those numbers. v is right there. g is, of course, 9.8. And if you double-check those numbers yourself, you should get 2.76 meters. Now, is that a realistic number to be able to jump? No. 2.76 meters, that's like 10 feet. Okay. So it's a little bit extreme, but the approach is sound. So hopefully that problem is clear. If you have any questions, come see me in my office. Cheers.