Physics

A stone is thrown straight up with an initial speed of 80 ft/s. How high does the stone go, and how long does it stay in the air?

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asked by Addie
  1. It reaches a height H such that
    M g H = (1/2) M Vo^2, therefore
    H = Vo^2/(2g)

    g = 32.2 ft/s^2 Vo is the initial velocity

    H = 99.4 feet

    Time going up = Vo/g = 2.48 s
    The time coming down is the same.

    Total time in the air = 4.96 s

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    posted by drwls

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