Question

A basketball player, standing near the basket to grab a rebound, jumps 65.5 cm vertically. How much time does the player spend in the top 13.6 cm of his jump? How much time does the player spend in the bottom 13.6 cm of the jump? Did you take into account the player jumping up AND falling down?

Answer 1

initial velocity of jump is given by

u^2 = 2 gh

u^2 = 2*9.8* 0.655

u = 3.583 m/s

now using 2nd equation of motion to find time

h = ut - 0.5 gt^2

0.136 = 3.583 t - 4.9 t^2

solving for t

t = 0.04 s ( during rise)

t = 0.691 s ( during fall)

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