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17 March, 10:55

A ball is thrown horizontally from the top of a tall cliff. Neglecting air drag, what vertical distance will the ball have fallen after 3 seconds?

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  1. 17 March, 12:54
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    The relevant equation to use here is:

    y = v0 t + 0.5 g t^2

    where y is the vertical distance, v0 is initial velocity = 0, t is time, g = 9.8 m/s^2

    y = 0 + 0.5 * 9.8 * 3^2

    y = 44.1 meters
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