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.] A. Calculate the impulse experienced by the person. B.

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. C. 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

From

v^2 = vo^2 - 2gh

Here, h = 2.8, vo = 0 m/s. Thus,

v = 7.408 m/s   [just before hitting the ground]

Thus, the impulse of hitting the ground is

pf - pi = 0 - 55 kg * 7.408 m/s

= -407 kg* m/s   [ANSWER, FIRST PART]

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It is initally 7.408 m/s before hitting the ground. Using

v^2 = vo^2 + 2ad

where

v = 0, vo = 7.408 m/s, d = 1 cm = 0.01 m,

a = (v^2 - vo^2)/2d

a = 2744 m/s^2

Thus,

ma = m(v^2 - vo^2)/2d = the net force applied

As

ma = N - mg

---> N = mg + ma

Normal force = 1.51E5 N   [ANSWER, SECOND PART]

******************

If it moves 50 cm instead of 0.01 m,

a = (v^2 - vo^2)/2d

a = 54.878 m/s^2

Thus,

---> N = mg + ma

Normal force = 3.56E3 N   [ANSWER, THIRD PART]

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