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A ball is thrown from a height of 10m with certain initial downward velocity. It hits the ground and loses 50% of its energy during the impact and rises back to the sane height. What was its initial downward velocity of projection?

  • science -

    initial energy
    = (1/2)m v^2 + m g (10)
    = m(.5 v^2+98.1)

    final energy = m(98.1)
    so
    98.1 = .5(.5v^2 +98.1)
    2(98.1) = .5 v^2 + 98.1
    .5 v^2 = 98.1
    v = sqrt(2*98.1)
    v = about 14 m/s

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