# Newton's laws and jumping

1. Jul 8, 2006

### Tensaiga

Ok here's my question:
An exceptional vertical jump from rest would raise a person 0.80m off the ground. to do this, what constant force would a 70kg person have to exert against the ground? assuming the person lowers himself by 0.2m prior to jumping and remains ina standing position while in the air.

ANS-the correct answer says in our book is 3.4 * 10^3 N. I didn't get that, i got something like 1300N.

the initial velocity i found was 3.43m/s, and i know it must over come the weight (gravtional field strength), and THAT's all i know...

Thanks

2. Jul 8, 2006

### nrqed

How did you get 3.43 m/s? I get something different.

How did you get 1300 N?
You must first find the acceleration of the person as he is pushing himself against the ground. what did you get?

EDIT: Btw, I get the same answer as the book.

Patrick

Last edited: Jul 8, 2006
3. Jul 9, 2006

### Tensaiga

what...i must have done something wrong, ok i used Vf^2 - Vi^2 = 2ad, (d=0.8-0.2=0.6m), Vi = rest, 0m/s, Vf = (that's what i'm trying to find), then deceleration due to gravity, 9.8m/s^2 downwards, and you plug it in it gives you 3.43m/s, then i used that to find the time which is 0.35s (d = (v1 + v2)/2 * t), then i found out accerlation funny, it's 9.8 again, then i used to to find Fnet, and finally plus fg. and i got 1370N

4. Jul 9, 2006

### tony873004

wouldn't you want to add .8 to .2 for a total distance of 1 meter?

5. Jul 9, 2006

### Tensaiga

but then what would you do after that?

6. Jul 9, 2006

### arildno

Tensaiga:
You are going down a wrong track here!

First, determine the jumper's initial velocity just when he leaves ground.
Use energy conservation for this.

Secondly, find the jumper's average acceleration from the time he has his legs lowered till he leaves ground.

Thirdly, determine the average force he's exerting on the ground.

7. Jul 9, 2006

### nrqed

You are going about it the wrong way, I am afraid. Arildno already explained how to do it, I will just add a few comments.

You must decompose the motion in two parts.

The first part, the person pushes against the ground with his feet. The displacement is 0.2 meter.

The second part, the person is in free fall. There the displacement is 0.8 m (not 0.6 m). Then the only force acting is the gravitational force.

Start from the second part to find the initial velocity just as the person is leaving the ground (that's using the equations you used except that the distance is 0.8 m. No need to find the time! And of course the acceleration is simply 9.8 m/s^2 downward)

Now, from knowing the final velocity when he leaves the ground, go back to part a. For that part, the initial velocity is zero, the final velocity is what you have just found in part b, you need to find the acceleration during that part (which is NOT 9.8 m/s^2 since the weight is not the only force). Once you find the acceleration, set up a free body diagram, find the normal force exterted by the ground on the person. From the action-reaction principle, this is also the force exerted by the person on the ground. You are done.

Hope this helps

Patrick

8. Jul 10, 2006

### Tensaiga

i see i see thank you every much

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