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- \frac{g*t^2}{2} $

 yo=0
y=120 cm =1,2m

$v_f^2 =v_o^2-2g*y $

$v_o= \sqrt{2g*y} = \sqrt{2*9,8*1,2} = 4,849m/s$
$v_f=v_o -gt$

$t = \frac{v_o}{g} =0,49s$

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