When jumping straight down, you can be seriously injured if you land stiff-legge

Question

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 72.4-kg man just before contact with the ground has a speed of 6.31 m/s. (a) In a stiff-legged landing he comes to a halt in 4.64 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 0.222 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).

Explanation / Answer

We will apply the impulse momentum theorem to solve this problem. From this theorem we know that, for a given change in momentum, greater forces are associated with shorter time intervals. Therefore, we expect that the force in the stiff-legged case will be greater than in the knees-bent case.

a) Assuming that upward is the positive direction, we find from the impulse-momentum theorem that

(F) = (mvf - mv0)/delta t = (72.4*0 - 72.4*(-6.31))/4.64*10^-3 = 98457.76 N

b) Again using the impulse-momentum theorem, we find that

  (F) = (mvf - mv0)/delta t = (72.4*0 - 72.4*(-6.31))/0.222 = 2057.86 N

c) The net average force acting on the man is (F) = Fground + W, where FGround is the average upward force exerted on the man by the ground and W is the downwardacting weight of the man. It follows, then, that

FGround = (F) - W. Since the weight is W = mg, we have

Stiff legged FGround = (F) - W =  98457.76 - [-72.4*9.8] = 99167.28 N

Knees bent   FGround = (F) - W = 2057.86 - [-72.4*9.8] = 2767.38 N

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