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16 November, 05:25

A 0.54-kg ball is thrown vertically upward with an initial speed of 15.5 m/s. What is the potential energy of the ball when it has travelled one-quarter of the distance to its maximum height

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  1. 16 November, 09:06
    0
    16.2J

    Explanation:

    At the maximum height v = 0

    V ^2 = u^2 - 2gh ( - because since the ball is going upward)

    h = u^2 - v^2 / 2g

    15.5^2 / 20

    h = 12.0125

    h = 12m

    At one quarter of the distance 1/4 * 12 = 3m

    Potential energy = mgh

    = 0.54 * 10 * 3

    = 16.2J
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