from a cliff of 49m high ,a man drops a stone 1s later he throes second stone . they both hit the ground at the same time . find out the speed with which he throws the second stone.

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asked by divya
  1. 1st Stone:
    4.9t^2 = 49
    t^2 = 10
    Tf = 3.16 s. = Fall time.

    2nd stone:
    h = Vo*t + 0.5g*t^2 = 49
    t = 3.16-1 = 2.16 s. = Fall time.

    Vo*2.16 + 4.9*2.16^2 = 49
    2.16Vo + 22.86 = 49
    2.16Vo = 49-22.86 = 26.14
    Vo = 12.1 m/s. = Initial velocity.

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    posted by Henry
  2. Nice

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    posted by Himani

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