Question

The best rebounder in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 120 cm. A: What is their initial "launch" speed off the ground? B: How long are they in the air?

Answer 1

`y=yo+vo*t- (g*t^2)/(2)`

yo=0
y=120 cm =1,2m


`v_f^2 =v_o^2-2g*y`


`v_o= √(2g*y) = √(2*9,8*1,2) = 4,849m/s`

`v_f=v_o -gt`


`t = (v_o)/(g) =0,49s`

t is the time to get just the maximum height.

so the total time he spent in air is Ttotal=2*t

Ttotal= 0,98s