Question

A basketball player, standing near the basket to grab a rebound, jumps 75.3 cm vertically. How much time does the player spewnd in the top 14.6 cm of his jump? How much time does the player spend in the bottom 14.6 cm of the jump?

Answer 1

a) we can answer the first part of this by recognizing the player rises 0.753 m, reaches the apex of motion, and then falls back to the ground

we can ask how long it takes to fall 0.146 m from rest:

dist = 1/2 gt^2 or t=sqrt[2d/g]
t=0.172 s

this is the time to fall from the top; it would take the same time to travel upward the final 0.146 m, so the total time spent in the upper 0.146 m is

2x0.172 = 0.344s

b) there are a couple of ways of finding the time it takes to travel the bottom 0.146m

we can use

d=1/2gt^2 twice to solve this problem

the time it takes to fall the final 0.146 m is:

time it takes to fall 0.753 m - time it takes to fall 0.607 m

t = sqrt[2d/g] = 0.392 s to fall 0.753 m, and

this equation yields it takes 0.352 s to fall 0.63 m, so it takes 0.04 s to fall the final 0.146 m. The total time spent in the lower 0.146 m is then twice this, or 0.08s