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If a pro basketball player has a vertical leap of about 35 inches, what is his hang time? Use the hang-time function V = 48T²

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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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