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 (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 110.0 kg.

Answer 1

Answer:

4.20214231077 m/s

29.429 m/s²

4316.29 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of player = 110 kg

Equations of linear motion

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 0.9}\\\Rightarrow u=4.20214231077\ m/s$

The velocity is 4.20214231077 m/s

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{4.20214231077^2-0^2}{2\times 0.3}\\\Rightarrow a=29.429\ m/s^2$

The acceleration is 29.429 m/s²

Force is given by

$F=m(a+g)\\\Rightarrow F=110(29.429+9.81)\\\Rightarrow F=4316.29\ N$

The force he exerts is 4316.29 N