# Impulse-Momentum Theorem

1. Oct 29, 2005

### chipsdeluxe

When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 66.9-kg man just before contact with the ground has a speed of 4.22 m/s. (a) In a stiff-legged landing he comes to a halt in 4.11 ms. Find the magnitude of the average net force that acts on him during this time. (b) When he bends his knees, he comes to a halt in 0.278 s. Find the magnitude of the average net force now. (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of the forces, find the magnitude of the force applied by the ground on the man in part (b).

Please help. this is how i solved the problem which is wrong:

a. Mman=66.9 kg
Vman=4.22 m/s
T=4.11 ms (1/1000)=0.00411 s
i used the equation:
F=-(m/t)v= -(66.9/0.00411)(4.22)=-68690.5 N
b. i used the same equation and only changed the time
c. i don't even know where to begin and i can't solve this until i get part a and b right.

thanks in advance

2. Oct 30, 2005

### daniel_i_l

In these kind of questions you should use the fact that the impulse J equals:
J = avgF*delta t. and J also equals the change in momentum J = m(Vf-Vi)
with those two equations you should be able to solve it.
I'm not sure what equation you were using?

3. Oct 30, 2005

### Andrew Mason

Your approach is correct.
The net force provides the acceleration:
$$F_{avg}=ma_{avg} = m\frac{\Delta v}{\Delta t}$$
Your answer for a. appears to be correct.

But this is not the only force on the body. We are ignoring gravity which is acting (downward) at all times. But because it is balanced by the ground, it does not result in any downward acceleration. So the vector sum of all forces is:

$$m\vec g + \vec N = m\vec{a_{avg}}$$

Solve for N.

AM

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