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23 August, 08:05

A ball is dropped from somewhere above a window that is 2.00 m in height. As it falls, it is visible to a person boxing through the window for 200 ms as it passes by the 2.00 m height of the window. From what height above the top of the window was the ball dropped?

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  1. 23 August, 10:14
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    4.14 m

    Explanation:

    In the last leg of the journey the ball covers 2 m in 2ms or 0.2 s.

    Let in this last leg, u be the initial velocity.

    s = ut + 1/2 g t²

    2 =.2 u +.5 x 9.8 x. 04

    u = 9.02 m / s.

    Let v be the final velocity in this leg

    v² = u² + 2 g s

    v² = (9.02) ² + 2 x 9.8 x 2

    = 81.36 + 39.2

    v = 10.97 m / s

    Now consider the whole height from where the ball dropped. Let it be h.

    Initial velocity u = 0

    v² = u² + 2gh

    (10.97) ² = 2 x 9.8 h

    h = 6.14 m

    Height from window

    = 6.14 - 2m

    = 4.14 m
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