Physics..............help please?

Question

A basketball player grabbing a rebound jumps 73.9 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top 10.8 cm of this jump and (b) in the bottom 10.8 cm? Do your results explain why such players seem to hang in the air at the top of a jump?

Answer 1

The ascent is identical to the descent, so we can just double the time. let's just look @ the descent.

On the descent, the player starts at y = 0.739m with velocity = 0 (because he's at the top of his height, where the velocity is at that moment, 0 m/s)  and ends at y = 0 so this is just free fall.

Consider first the top 0.108m of the jump:

y = 1/2 g t^2

-.108 = 1/2 * -9.81 t²

t = 0.148s

Now consider the bottom 0.108m.

Time to hit y =0 (ground)

y = -1/2*g*t² ;  y = -0.739m

t = 0.388s

Notice that y = negative here because we're traveling downward.

We start at 0.739m and end at 0.108 m, which is a difference of 0.631 m

y = 1/2 g t²

t = .359s

Subtract these to get the net time = 0.0295s

Now, we can just double these times since the time to ascend to that height = the time to descend

(a) 0.297 s

(b) 0.0590 s