# Impulse Momentum Theorem

1. Jul 6, 2007

Impulse Momentum Theorem!!

1. The problem statement, all variables and given/known data
A student (m = 63 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.0180 s. The average force exerted on him by the ground is +16000. N, where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.

2. Relevant equations

impulse momentum theorem

3. The attempt at a solution

i did it and got vfinal=4.57 and put that over the time, 0.018sec and got 254 m

but thats not right.... i dont know why...anyone able to help?

2. Jul 6, 2007

### Staff: Mentor

Did you include the force of gravity?
No idea what you did here. If something fell from that height, how fast would it be going?

3. Jul 6, 2007

I'm a little lost on how to approach this problem :(

4. Jul 6, 2007

### Staff: Mentor

Once you've found the speed, finding the height is just a falling body problem.

5. Jul 6, 2007

Ok, so can I just use F=ma and say 16000 N=63kg x (a) and get acceleration and then use kinematics to say

displacement=1/2 (a) (0.018)sqrd

?

6. Jul 6, 2007

### Staff: Mentor

No. You are finding the height the student fell from to reach the speed that you figured out. (You do not need the displacement during his interaction with the ground.) The acceleration is the acceleration due to gravity.

How did you find the speed?

7. Jul 6, 2007

I made 16000=63v/.018 and got 4.57 :/

8. Jul 6, 2007

### Staff: Mentor

That's the impulse momentum theorem--good. But realize that you need the net force, not just the force from the ground. What other force acts on the student? (This correction will give you a slightly different speed.)

9. Jul 6, 2007

Ah, weight=mg=617.4

so i get v=4.395

10. Jul 6, 2007

what could i do next?

11. Jul 6, 2007

Oh, in the problem it says : "Assume that the only force acting on him during the collision is that due to the ground."

so that's all i have to worry about. i have the right velocity as 4.57 m/s then...i dont know what to do next however.

12. Jul 6, 2007

### Staff: Mentor

Now figure out the height he must have fallen from. This part has nothing to do with the collision with the ground or with the impulse-momentum theorem.

13. Jul 6, 2007

I keep getting really small answers. Like, wouldn't I just do 4.57m/s=x/.018sec?

14. Jul 6, 2007

### Staff: Mentor

This is not correct. The 0.018 sec is the duration of the impact with the ground, not the time it takes him to fall. (Also: Falling is accelerated motion, not constant speed.)

Review your basic kinematic relationships. (Hint: You can also use energy methods, if you like.)

15. Jul 6, 2007

Ok, I did it again and got

y=(0)-(4.57 squared)/(2)(-9.8)

and i got -1.066m

would that be what you got?

16. Jul 6, 2007