# Can anyone answer this?

1. Oct 13, 2007

### AcroNinja

I was watching a show on TV that showed how many lbs. of force a basketball players uses to jump. In this case, the player pushed into the floor with 1400 lbs. of pressure to jump 4 feet in the air. That's all good and well, but does this mean when the player comes down and lands he will impact with 1400 lbs.? So if I were to jump with, say, 1000lbs. of pressure, do I land with 1000 lbs. of impact into the floor? Just curious, thanks everyone. In other words, is the jump the same as the impact?

2. Oct 13, 2007

### notknowing

Hi,
Good question. I'm used thinking in terms of Newton and meter but that is not essential here. The force at impact is not necessarily the same as the force during the jump, though it can be. The force during the jumping phase is at each moment equal to mass times acceleration and this acceleration (or deceleration) is not necessarily the same during jump and impact. During impact, one could decelerate over a very short time (by holding the legs as stif as possible) or one could decelerate over a longer time (by going deep through the knees) and both cases result in different forces (exerted over different distances). The only thing one can say is that the energy generated during the jumping phase is roughly equal to the energy released at impact. Hope this answers your question.

Rudi Van Nieuwenhove (Norway)

3. Oct 13, 2007

### AcroNinja

It kind of answers my question. You never actually said the impact is the same as the takeoff. I just need to hear that clearly. That's what I am looking for, or looking to have something disagree with. Is the force of lbs. to jump off the ground the same as it is when you come back down and make an impact? Are they equal?

4. Oct 13, 2007

### Staff: Mentor

He said the energy is the same. The force may or may not be the same, depending on how you land. If you let your knees bend exactly as they did when you jumped, the force will be the same.

5. Oct 13, 2007

### nrqed

What is the same just after takeoff and at just before landing is the speed (neglecting air friction). So the magnitude of the momentum is the same. Let's call this value "p". The force is a change of momentum over a time interval, $F = \frac{\Delta p}{\Delta t}$ (I am being a bit careless here with magnitude vs components).

At the takeoff and at the landing the change of momentum is the same. However, the time intereval during which the momentum is imparted/dissipated may be very different. For example, if the player bends the knees a lot while landing, the change of momentum will be done over a long time interval. Same thing while jumping.

Last edited: Oct 13, 2007
6. Oct 13, 2007

### sneez

It does not have to be or it could be. It has to do with the concept of impulse. That is proportional to the force and time interval over which it acts. So if for example one bend his knees during landing , hence extending the time over which the change in momentum takes place, hence reducing the impact force.

7. Oct 13, 2007

### AcroNinja

I understand impulse, maybe I should have said the player does not bend their knees when landing. Even if they do bend their knees, it's still the same weight right? It's just being absorbed and distributed better though your legs . Let me use a different example, one with out muscles. If you had a machine that shot a baseball up 100 feet in the air from ground level, and you let the ball come back down 100 feet and land on your face as you are laying on the ground, would this be the same impact as if you stuck your face right in front of the machine and let it hit you in the face point blank? Thanks everyone for putting up with me.

8. Oct 14, 2007

### keltix

force=mass*acceleration

The velocity with which the player took off should be the same as when he lands. But I don't know about acceleration. If acceleration is the same then theoretically the force should be the same (if mass maintains).

9. Dec 16, 2007

### vector3

A lot of good responses preceed this but, I believe this is a "spring-mass" problem. The spring being the players legs and the mass being the players body. The spring constant, k, of the players legs being a variable determined by the player during the actual playing conditions. In otherwords, he may launch quickly to get to 4 feet height but may or may not absorb the free fall energy in his legs in the same way. Depends on the playing conditions. In the most technical respects, the energy to launch will drive the player to potential energy state $mgh$ however the return direction the force of drag will do work against the player and so the energy for his legs to absorb will be less.

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