A little more physics help looking for time?

Question

A basketball player grabbing a rebound jumps 70.6 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top 10.5 cm of this jump and (b) in the bottom 10.5 cm? Do your results explain why such players seem to hang in the air at the top of a jump?

Answer 1

I think the easiest way to solve this is to start by considering the player at the top of his/her jump.  At this point, the player is 70.6 or 0.706 m above the ground and has zero vertical speed.  In essence, the person is falling from rest from a height of 0.706 m.

Let's calculate how long it takes to fall 10.5 cm, 70.6 cm, and 60.1 cm (which is how far you have to fall to be 10.5 cm above the ground)

We use the same equation for each:

distance = 1/2 g t^2 where g is the accel due to gravity (9.8 m/s/s) and we solve for t

it takes 0.1464 s to fall 10.5 cm or 0.105 m

it takes 0.3796 s to fall 70.6 cm or 0.706m

it takes 0.3531 to fall 61.1 cm, or in other words, the athlete takes

0.3796-0.3531=0.0265s to fall the last 10.5 cm...so it does seem as if this explains why athletes seem to "hang" in the air

(another reason is that athletes also change their body position like ballet dancers...but this is for when you study moment of inertia)