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15 January, 09:58

A 0.5 kg ball moving at a speed of 3m/s rolls up a hill. How high does the ball roll before it stops

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  1. 15 January, 12:03
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    According to conservation of energy

    Kinetic energy at the bottom of hill = potential energy at top of hill

    Kinetic energy (KE) = 0.5*mass (m) * velocity (v) ²

    where m=0.5 kg, and v = 3 m/s, v²=9

    KE=0.5*0.5*9=2.25 J = PE

    PE=m*gravitational acceleration (g) * height (h)

    h=2.25 / (0.5*9.8) = 0.459 m
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