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1 May, 23:50

A basketball player can leap upward 0.43 m. how long does he remain in the air? use an acceleration due to gravity of 9.80 m/s2 and find your answer to the nearest 0.001 s.

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  1. 1 May, 23:56
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    From the equations of linear motion,

    v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,

    Thus, v² = u² - 2gs, but v=0

    hence, u² = 2gs

    = 2*9.81*0.43

    = 8.4366

    u = √8.4366

    =2.905 m/s

    Hence the initial velocity is 2.905 m/s

    Then using the equation v = u + gt.

    Therefore, v = u - gt. (-g because the player is jumping against the gravity)

    but, v = 0

    Thus, u = gt

    Hence, t = u/g

    = 2.905/9.81

    = 0.296 seconds
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