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21 September, 10:56

If michael jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time

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  1. 21 September, 11:07
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    The formula for the vertical height, h, attained with a vertical take off speed of u is

    u² - 2gh = 0

    where

    g = 9.8 m/s², acceleration due to gravity.

    Because h = 1.29 m, therefore

    u² = 2*9.8*1.39 = 27.244

    u = 5.2196 m/s

    Also, the time of flight, t, required when the vertical take-off speed is u is given by

    ut - 0.5gt² = 0.

    Therefore

    5.2196t - 0.5*9.8*t² = 0

    5.2196t - 4.9t² = 0

    t (5.2196 - 4.9t) = 0

    t = 0, or t = 5.2196/4.9 = 1.0652 s

    t = 0 corresponds to take-off.

    t = 1.0652 s corresponds to landing.

    The hang time is 1.0652 s.

    Answer:

    Take-off speed = 5.22 m/s (nearest hundredth)

    Hang time = 1.065 s (nearest thousandth)
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