Physic C _Help me plz

Question

An 80kg man jumps down to a concrete patio from a window ledge only 0.50 m above the ground. He neglects to bend his knees on landing, so that his motion is arrested in a distance of 2.0cm.

a) What is the average acceleration of the man from the time his feet first touch the patio to the time he is brought fully to rest?

b) With what force does this jump jar his bone structure?

Answer 1

a)

From potential energy to kinetic energy,

mgh = (1/2)mv^2

2gh = v^2

Assuming constant deceleration over the 2 cm,

v^2 - 0 = 2as

so

2gh = 2as

a = gh/s

a ≈ (0.50 m)(9.80665 m/s^2)/(0.02 m)

a ≈ 245.16625 m/s^2 ≈ 245 m/s^2, or 25 g's

b) F = ma

F ≈ (80 kg)(245.16625 m/s^2)

F ≈ 19,613.3 N ≈ 19,600 N

This is why bones get broken from seemingly trivial falls and accidents.