Hang time at certain points for someone jumping straight up...?

Question

A basketball player, standing near the basket to grab a rebound, jumps 76.0 cm vertically. How much total time does the player spend (a) in the top 15.0 cm of this jump and (b) in the bottom 15.0 cm? Does this help explain why such players seem to hang in the air at the tops of their jumps? Show and explain your work. Thanks and good luck! :)

Answer 1

Half the total time in the air is calculated from
.76m=.5*9.81*t^2
t=sqrt(2*.76/9.81)
0.39 seconds
so the total time is
0.78 seconds

The speed at take-off is
0.39*9.81
3.83 m/s

So the time in each of the bottom 15 cm is
.15=3.83*t-.5*9.81*t^2
compute the smaller root for t
0.041 sec
or a total of 0.082 sec

Half the time at the top is
.15=.5*9.81*t^2
t=sqrt(0.30/9.81)
0.17 seconds
or a total of 0.34 seconds, more than four times longer - hence the "hang"

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