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A basketball player standing near the basket to grap a rebound, jumps 62 cm vertically. How much total time does the player speen in the top 15 cm of the jump? How much total time does the player spend in the bottm 15 cm of this jump?

  • physics -

    If he jumps up H = 0.62 m, the initial velocity V is sucn that
    (1/2) M V^2 = M g H
    V = sqrt (2gH) = 3.48 m/s

    Height vs time is given by
    y = 3.38 t - 4.90 t^2
    Solve for the times when H = 0.15 m and y = 0.47 m. You will need to use the postive root of the quadratic eauation.


    At the top of the trajectory, H = 0.62 m and g t = V = 3.48 m/s, so t = 0.355 s there.
    t = 0 when y = 0. Knowing the times when H = 0.15 and 0.47 m will let you calculate the intervals they are asking for.

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