Physics

A ball is thrown directly upward with an initial velocity of 14 m/s. If the ball is released from an initial height of 2.8 m above ground, how long is the ball in the air before landing on the ground? Ignore air drag.

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  1. Vf = Vo^2 + 2gd,
    d(up) = (Vf^2 - Vo^2) / 2g,
    d(up) = (0 - (14)^2) / -19.6 = 10m.

    D = 2.8 + 10 = 12.8m above ground.

    Vf = Vo + gt,
    t(up) = (Vf - Vo) / g,
    t(up) = (0 - 14) / -9.8 = 1.43s.

    D = Vo*t + 0.5gt^2 = 12.8,
    0 + 0.5*9.8t^2 = 12.8,
    4.9t^2 = 12.8,
    t^2 = 2.61,
    t(down) = 1.62s.

    T = 1.43 + 1.62 = 3.05s. in air.

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