A 75.0-kg man steps off a platform 3.10 m above the ground. he keeps his legs straight as he falls, but his knees begin to bend at the moment his feet touch the ground; treated as a particle, he moves an additional 0.60 m before coming to rest. (a) what is his speed at the instant his feet touch the ground? (b) if we treat the man as a particle, what is his acceleration (magnitude and direction) as he slows down, if the acceleration is assumed to be constant?

Refer to the diagram shown below.

u = 0, the initial vertical velocity

Assume g = 9.8 m/s² and ignore air resistance.

At the first stage of landing on the ground, the distance traveled is

h = 3.1 - 0.6 = 2.5 m.

If v = the vertical velocity at this stage, then

v² = u² + 2gh

v² = 2 * (9.8 m/s²) * (2.5 m) = 49 (m/s) ²

v = 7 m/s

At the second stage of landing on the ground, let a = the acceleration (actually deceleration) that his body provides to come to rest.

The distance traveled is 0.6 m.

Therefore

0 = (7 m/s) ² + 2 (a m/s²) * (0.6 m)

a = - 49/1.2 = - 40.833 m/s²

Answers:

(a) The velocity when the man first touches the ground is 7.0 m/s.

(b) The acceleration is - 40.83 m/s² (deceleration of 40.83 m/s²) to come to rest within 0.6 m.