Physics

A ball is thrown straight upward and rises to a maximum height of 20 m above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value?

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asked by Cara
  1. v = Vi - g t
    at max height, v = 0
    0 = Vi -gt
    so t at top = Vi/g

    How high does it go?
    20 = Vi t - (1/2)g t^2
    20 = Vi (Vi/g) - (1/2) gVi^2/g^2

    20 = (1/2) Vi^2/g
    40 (9.81) = Vi^2
    Vi = 19.8 m/s

    when does v = 19.8/2 or 9.9 m/s
    v = Vi - g t
    9.9 = 19.8 - 9.81 t
    9.81 t = 9.9
    t = 1 second

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    posted by Damon

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