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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 72.7-kg man just before contact with the ground has a speed of 8.46 m/s. (a) In a stiff-legged landing he comes to a halt in 4.10 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 0.166 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).

1 Answer

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(a)
1.5\cdot 10^5 N

According to the impulse theorem, the impulse exerted on the man during the impact is equal to his change in momentum:


I=\Delta p\\F \Delta t = m \Delta v

where we have

F = magnitude of the average force


\Delta t=4.10 ms = 0.0041 s is the contact time


m=72.7 kg is the mass of the man


\Delta v = v_f -v_i = 0-(8.46 m/s)=-8.46 m/s is the change in velocity of the man

Solving the formula for F, we find


F=(m \Delta v)/(\Delta t)=((72.7 kg)(-8.46 m/s))/(0.0041 s)=-1.5\cdot 10^5 N

And the negative sign simply means the direction of the force is opposite to the initial velocity of the man (so, the force points upward).

(b) 3705 N

This part of the exercise is exactly identical to part (a), but here the contact time is much longer:


\Delta t=0.166 s

Substituting into the equation, we find


F=(m \Delta v)/(\Delta t)=((72.7 kg)(-8.46 m/s))/(0.166 s)=-3705 N

(c) 2992.5 N

We have two forces acting on the man:

- The force that the ground exerts on the man, pointing upward:


N=3705 N

- The force of gravity (weight of the man), pointing downward:


W=mg=(72.7 kg)(9.8 m/s^2)=712.5 N

Since the directions are opposite, the resultant force will be


F=N-W=3705 N-712.5 N=2992.5 N

and the direction is upward.

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