Basketball hangtime physics problem

AI Thread Summary
The discussion focuses on calculating the total time a basketball player spends in the top and bottom 13.3 cm of a 75.4 cm vertical jump. For part (a), the correct ascent time was determined to be 0.3325 seconds using the equations vf^2=vi^2-2ay and d=v/t. However, part (b) posed challenges, as the attempt to calculate the descent time using dy=vit-.5gt^2 resulted in an incorrect answer. It was noted that the total time for both ascent and descent needs to be considered for accurate results. The conversation highlights the physics behind the perception of players "hanging" in the air during jumps.
wbetting
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Homework Statement


A basketball player grabbing a rebound jumps 75.4 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top 13.3 cm of this jump and (b) in the bottom 13.3 cm? Do your results explain why such players seem to hang in the air at the top of a jump?


Homework Equations



vf^2=vi^2-2ay
d=v/t

The Attempt at a Solution


i got part a correct by using vf^2=vi^2-2ay to find 1.6 m/s then i used d=v/t which i rewrote as t=2y/vi+vf and got .3325 for my answer which was correct. then part b came and i tried to use dy=vit-.5gt^2 and i got .0767 which was wrong
 
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Hi wbetting!

wbetting said:

The Attempt at a Solution


i got part a correct by using vf^2=vi^2-2ay to find 1.6 m/s then i used d=v/t which i rewrote as t=2y/vi+vf and got .3325 for my answer which was correct. then part b came and i tried to use dy=vit-.5gt^2 and i got .0767 which was wrong

You need to find the total time the basketball player is in both ascent and descent. Your equation will only give you one of those. :wink:
 
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