# Constant acceleration problem

1. Sep 14, 2010

### apiwowar

A basketball player grabbing a rebound jumps 76.1 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top 14.4 cm of this jump and (b) in the bottom 14.4 cm? Do your results explain why such players seem to hang in the air at the top of a jump?

so the only thing i thought of doing was finding the initial velocity

Vi^2 = -2ax
Vi= 3.86 m/s

now i have no idea what do to.
any help?

2. Sep 14, 2010

### Delphi51

You need to find the time the jumper reaches height 76.1-14.4.
So you need a distance as a function of time formula.

3. Sep 15, 2010

### iffydroplight

Up an down times are equal so begin by solving for descent data.
At 76.1cm, vertical velocity is zero.

Calculate velocity after falling 14.4cm. This can be plugged into vi in the time equation to determine seconds falling.

Begin at 0. Solve for -14.4cm. Apply to your time equation. Time 1 - time 0 = 'hang time'/2.

Do the same for -(76.1cm-14.4cm) to find time at 14.4 cm above the gym floor. Do the same for -76.1cm. Subtract again.

Make sure you reconcile up plus down times (multiply by two).