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September 16, 2014

September 16, 2014

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

- math -
**MathMate**, Sunday, April 24, 2011 at 3:06pmDistance, 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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