Question

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 100 cm . a) What is their initial "launch" speed off the ground?

b)How long are they in the air?

Answer 1

a) 4.45 m/s

b) 0.9 seconds

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

`v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=√(v^2-2as)\\\Rightarrow u=√(0^2-2* -9.81* 1)\\\Rightarrow u=4.45\ m/s`

a) The vertical speed when the player leaves the ground is 4.45 m/s

`v=u+at\\\Rightarrow t=(v-u)/(a)\\\Rightarrow t=(0-4.45)/(-9.81)\\\Rightarrow t=0.45\ s`

Time taken to reach the maximum height is 0.45 seconds

`s=ut+(1)/(2)at^2\\\Rightarrow 1=0t+(1)/(2)* 9.81* t^2\\\Rightarrow t=\sqrt{(1* 2)/(9.81)}\\\Rightarrow t=0.45\ s`

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds