Posted by **Juan** on Sunday, April 24, 2011 at 1:16pm.

If a pro basketball player has a vertical leap of about 35 inches, what is his hang time? Use the hang-time function V = 48Tē

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**MathMate**, Sunday, April 24, 2011 at 3:06pm
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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