# Weight of landing from jumping

1. Nov 2, 2007

### andrewz182

if i were a 160 lb person who could jump a height of 2 feet, what would be my "weight/force/momentum" or whatever measurement used that will result in a calculation of pounds in the end.

this is in no way me asking for help with hw, this is over a debate i had with a friend, and neither of us can really figure it out...

would it be (mass x acceleration x time)??

Last edited: Nov 2, 2007
2. Nov 2, 2007

### mjsd

what do you mean by
?

3. Nov 2, 2007

### andrewz182

or "pounds of force"

4. Nov 2, 2007

### mjsd

what do mean by "in the end"? I don't understand what is the debate about either

5. Nov 2, 2007

### andrewz182

i mean at the point of impact with the ground.

6. Nov 2, 2007

### andrewz182

ok to specify: what weight would a scale read if i were to jump on it from 2 feet?? would it just be calculating momentum??

7. Nov 2, 2007

### mjsd

so I guess you are saying that whether the downward force measured by a scale on the ground would register a bigger reading if you jump onto the scale as oppose to standing still on it? what do you think?

8. Nov 2, 2007

### andrewz182

but more specifically, would it be possible for a 160 lb person, who is able to jump 2 feet up in the air, to exert enough force upon a scale to read over 400 lbs?

9. Nov 3, 2007

### rcgldr

Force is equal to mass times acceleration. The same velocity can be achieved with a faster acceleration over a shorter period of time. There's no limit to the amount of force involved in a 2 foot landing, other than the maximum deceleration will be limited by how much compression there is at the contact point.

10. Nov 3, 2007

### mjsd

the force you are talking about is probably the instantaneous reading NOT at impact but when your body has just been fully decelerate to velocity zero I think, and before the scale re-adjust and show your "true weight"

mind you there are all those usual stuff like air resistance, or how you may destroy the scale etc...ie. what is the displacement during deceleration (otherwise it is probably not just 2 feet the total distance travelled) or whether deceleration is constant during that time....