Question

a basketball player can leap upward .65m how long does the basketball player remain in the air use 9.81m/s²​

Answer 1

At the player's maximum height, their velocity is 0. Recall that

`{v_f}^2-{v_i}^2=2a\Delta y`

which tells us the player's initial velocity
`v_i` is

`0^2-{v_i}^2=-2g(0.65\,\mathrm m)\implies v_i=3.6(\rm m)/(\rm s)`

The player's height at time
`t` is given by

`y=v_it-\frac g2t^2`

so we find their airtime to be

`0.65\,\mathrm m=\left(3.6(\rm m)/(\rm s)\right)t-\frac g2t^2\implies t=0.36\,\mathrm s`