Question

A basketball player, standing near the basket to grab a rebound, jumps 67.1 cm vertically.
How much time does the player spend in the top 14.4 cm of his jump?

How much time does the player spend in the bottom 14.4 cmof the jump?

Does this help explain why such players seem to hang in the air at the top of their jumps?

Answer 1

we can ask how long it takes to fall 14.4cm=0.144 m from rest:

dist = 1/2 gt^2 or t=sqrt[2d/g]

t= sqrt (2*0.144/9.8) =0.1714 s

this is the time to fall from the top; it would take the same time to travel upward the final 0.144 m, so the total time spent in the upper 0.144 m is

2x0.1714= 0.3428s

We need to have an initial velocity off of the ground - that cannot be zero.

So the jumper reaches a maximum height of 67.1 cm.

v² = u² + 2as

0 = u² - 2*(9.8)*0.671

u = 3.6265 m/sec

then time spent (t) to reach 14.4cm
0.144 = 3.626*t - 4.9*t²

Quadratic equation ? t = 0.042 sec

this is the time to cover this distance going up; it will take the same time to cover this distance coming down, so the total time spent in the lower 0.144 m is 0.084s