A 55-kg person lands on firm ground after jumping from a height of 2.8 m. [Hint:

Question

A 55-kg person lands on firm ground after jumping from a height of 2.8 m. [Hint: The average net force on him, which is related to impulse, is the vector sum of gravity and the force exerted by the ground.] (Figure 1). Calculate the impulse experienced by the person. Estimate the average force exerted on the person's feet by the ground if the landing is stiff-legged. Assume the body moves 1.0 cm during impact. Estimate the average force exerted on the person's feet by the ground if the landing is with bent legs. Assume the body moves 50 cm during impact.

Explanation / Answer

dP = mdV = 55*sqrt(2*9.8*2.8) = 407.4457019 = 407 kg.m/s from the impact. ANS A.

The average deceleration on impact is given by

A = V^2/2S = (2*9.8*2.8)/(2*.01) = ? ;

so F = mA = 55*(2*9.8*2.8)/(2*.01) = 150920 N plus the weight

W = mg = 55*9.8 = ?.

So the normal force N = 151000 + 55*9.8 = 151539 = 152000 N. ANS B.

From B, we can modify and have N = 55*(2*9.8*2.8)/(2*.5) + 55*9.8 = 3557.4 = 3560 N. ANS C.

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