Question

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.

Answer 1

Answer:

a) Vo = 4.24m/s

b) a = 30m/s^2

c) F = 4400N

Explanation:

For part a) we study the movement after he leaves the ground. From that we know:

$Vf^2 = Vo^2 - 2*g*\Delta Y$  Where Vf=0  and ΔY=0.9m

Solving for Vo:

$Vo = \sqrt{2*g*\Delta Y} = 4.24m/s$

For part b) we have to study the movement before he leaves the ground.

$Vf^2 = Vo^2+2*a*\Delta Y$   Where Vo=0  and ΔY=0.3m

Solving for the acceleration:

$a=\frac{Vf^2}{2*\Delta Y}=30m/s^2$

For part c) we evaluate forces on the player

The force exerted by the floor on the player is the same as the one exerted by the player on the floor:

Fe - m*g = m*a     Fe = m*(g+a) = 4400N