A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0 m high cliff. At the instant the ball is thrown, a woman starts running away from the base of the cliff with a constant speed of 6.00 m/s. The woman runs in a straight line on level ground, and air resistance acting on the ball can be ignored.How far does the woman run before she catches the ball?


  1. Answer:

    The distance traveled by the woman is 34.1m



    The initial height of the cliff

    yo = 45m final, positition y = 0m bottom of the cliff

    y = yo + ut -1/2gt²

    u = 20.0m/s initial speed

    g = 9.80m/s²

    0 = 45.0 + 20×t –1/2×9.8×t²

    0 = 45 +20t –4.9t²

    Solving quadratically or by using a calculator,

    t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s

    So this is the total time it takes for the ball to reach the ground from the height it was thrown.

    The distance traveled by the woman is

    s = vt

    Given the speed of the woman v = 6.00m/s


    s = 6.00×5.69 = 34.14m

    Approximately 34.1m to 3 significant figures.

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