math
posted by Juan .
If a pro basketball player has a vertical leap of about 35 inches, what is his hang time? Use the hangtime function V = 48TÂ²

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Distance, V, in inches travelled from rest under gravity
= (1/2)gt²
= (1/2)32'*12"/' t²
Since the hang time T includes upward and downward movements, T=2t, so equation above becomes
192(T/2)² = V
hence
V=48T²
Substitute V=35" to get
35=48T²
Solve to get
T=sqrt(35/48)=0.854 sec.
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