SOLUTION: If a pro-basketball player has a vertical leap of 35 in, what is his hang time? Use the hang time function of V=48T^2. Hang time in seconds. Please help, thank you.

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Question 177861: If a pro-basketball player has a vertical leap of 35 in, what is his hang time? Use the hang time function of V=48T^2.
Hang time in seconds.
Please help, thank you.

Answer by nerdybill(7384)   (Show Source): You can put this solution on YOUR website!
Given:
V=48T^2
where
V is vertical leap
T is time (in seconds)
.
Plugging in the given vertical leap:
V=48T^2
35=48T^2
35/48 = T^2
0.72917 = T^2
Taking the square root of both sides:
0.854 secs = T

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