Question

Suppose you want to jump up first by crunching down and then pushing off the ground. You initially lower your center of mass by 0.2m. The average force exerted by the floor on you is three times your weight. What is your upward speed as you pass through your standing position just as you are leaving the floor?

Answer 1

Here,

for the net force acting on the person

Fnet = 3mg - mg = 2 * mg

Using work energy theorem

0.50 * m * v^2 = 2mg * 0.20

0.50 * v^2 = 2 * 0.20 * 9.8

solving for v

v = 2.8 m/s

the upwards speed as pass through the standing position is 2.8 m/s