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26 October, 03:24

A 180 pound person dives off of a 30 foot cliff. What is their velocity as they enter the water?

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  1. 26 October, 03:40
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    Taking the sea level as reference for altitude, then, on top of the cliff the diver has only potential energy:

    Ep=m*g*h, where m is the diver's mass, g=9.8m/s² and h=30ft=9.144m is the cliff's height.

    At the sea level, the diver has kinetic energy: Ec=m*v²/2

    From energy conservation: Ec=Ep, resulting v=sqrt (2*g*h) = 13.4m/s

    Answer: v=13.4m/s

    Note that mass does not count in the final result.
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