Question

You crouch from a standing position, lowering your center of mass 18 cm in the process. Then you jump vertically. The average force exerted on you by the floor while you jump is three times your weight. What is your upward speed as you pass through your standing position in leaving the floor?

Answer 1

To solve this problem we will apply the concepts related to balance. Since the force applied is 3 times the weight, and the weight is defined as the multiplication between mass and gravity, we will have that the dynamic equilibrium ratio would be given by the relation,

$\sum F = ma$

$3mg-mg = ma$

Rearranging to find a,

$a = 2g$

Using the linear motion kinematic equations, which express that the final velocity of the body, and in the absence of initial velocity, is equivalent to the product between 2 times the acceleration by the distance traveled, that is

$v^2 = 2as$

$v^2 = 2(2g)(0.18)$

$v^2 = 2(2*9.8)(0.18)$

$v = 2.66m/s$

Therefore the upward speed is 2.66m/s