Question

A 101 kg basketball player crouches down 0.380 m while waiting to jump. After exerting a force on the floor through this 0.380 m, his feet leave the floor and his center of gravity rises 0.920 m above its normal standing erect position. (a) Using energy considerations, calculate his velocity (in m/s) when he leaves the floor. m/s (b) What average force (in N) did he exert on the floor

Answer 1

`4.25\ \text{m/s}`

`3391.22\ \text{N}`

Step-by-step explanation:

y = Height of compression = 0.38 m

m = Mass of basketball player = 101 kg

h = Height of center of gravity after jump = 0.92 m

g = Acceleration due to gravity =
`9.81\ \text{m/s}^2`

Energy balance of the system is given by

`mgh=(1)/(2)mv^2\\\Rightarrow v=√(2gh)\\\Rightarrow v=√(2* 9.81* 0.92)\\\Rightarrow v=4.25\ \text{m/s}`

The velocity of the player when he leaves the floor is
`4.25\ \text{m/s}`

`Fy=mgy+(1)/(2)mv^2\\\Rightarrow F=(mgy+(1)/(2)mv^2)/(y)\\\Rightarrow F=(101* 9.81* 0.38+(1)/(2)* 101* 4.25^2)/(0.38)\\\Rightarrow F=3391.22\ \text{N}`

The force exerted on the floor is
`3391.22\ \text{N}`.