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25 March, 22:35

A 5.5 kg bowling ball initially at rest is dropped from the top of a 12 m building. It hits the ground 1.75 s later. Find the net external force on the falling ball.

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  1. 26 March, 02:11
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    We are given that:

    m = 5.5 kg

    d = 12 m

    t = 1.75 s

    Using the distance formula, we can get the acceleration of the ball:

    d = (1/2) a*t^2

    Rearranging:

    a = 2d/t^2 = 2 * (12 m) / (1.75 s) ^2

    a = 7.837 m/s^2

    Using Newton’s 2nd law, force is defines as:

    F = ma

    F = 5.5 kg (7.837 m/s^2)

    F = 43.10 N
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