physics

A boy drops a stone-A, down from a cliff 480 meters high, next he throws a stone-B, 1 second later - at what initial velocity was the stone-B be thrown so as to meet the stone-A on ground at the same time? g=10 m/s2.

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asked by alex
  1. Require the following:
    (g/2)T^2 = 480
    (g/2)(T-1)^2 + V*(T-1) = 480

    You have two equations in two unknowns (V and T) and can solve for V.

    The first equation tells you that stone A requires
    T = 9.8 seconds to fall to the ground. (if g= 10)

    Use that value of T in the second equation to solve for V.

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    posted by drwls

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