Question

A 67.5-kg basketball player jumps vertically and leaves the floor with a velocity of 1.99 m/s upward.

(a) What impulse does the player experience? magnitude 134.33

(b) What force does the floor exert on the player before the jump?

(c) What is the total average force exerted by the floor on the player if the player is in contact with the floor for 0.450 s during the jump?

Answer 1

-134.325 kg m/s

662.175 N

953.544 N

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Impulse

`J=m(v-u)\\\Rightarrow J=67.5(0-1.99)\\\Rightarrow J=-134.325\ kg m/s`

Impulse the player experiences is -134.325 kg m/s

Force

`F=mg\\\Rightarrow F=67.5* 9.81\\\Rightarrow F=662.175\ N`

The force does the floor exert on the player before the jump is 662.175 N

Equation of motion

`v=u+at\\\Rightarrow a=(v-u)/(t)\\\Rightarrow a=(0-1.99)/(0.45)\\\Rightarrow a=-4.422\ m/s^2`

`F=m(g-a)\\\Rightarrow F=67(9.81-(-4.422))\\\Rightarrow F=953.544\ N`

The force exerted by the floor on the player is 953.544 N