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a ball is thrown upward with an initial velocity of 38.9 m/s from the edge of a cliff of height H. The ball lands on the ground at the base of the cliff 12.8 s after it is thrown.

  • physics. -

    I assume you want H.

    h = H + Vi t + (1/2)(-9.8) t^2 = 0 at ground

    0 = H + 38.9 (12.8) - 4.9 (12.8)^2

    solve for H

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