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4 June, 02:29

4. Jimmy dropped a 10 kg bowling ball from a building that is 25 meters high.

Remembering that because of the Law of Conservation of Energy, PE I - KE, what

is the Kinetic Energy of the object right before it hits the ground?

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Answers (1)
  1. 4 June, 04:23
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    The K. E of the bowling ball right before it hits the ground, K. E = 2450 J

    Explanation:

    Given data,

    The mass of the bowling ball, m = 10 kg

    The height of the building, h = 25 m

    The total mechanical energy of the body is given by,

    E = P. E + K. E

    At height 'h' the P. E is maximum and the K. E is zero,

    According to the law of conservation of energy, the K. E at the ground before hitting the ground is equal to the P. E at 'h'

    Therefore, P. E at 'h'

    P. E = mgh

    = 10 x 9.8 x 25

    = 2450 J

    Hence, the K. E of the bowling ball right before it hits the ground, K. E = 2450 J
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