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Posted by on Wednesday, January 30, 2013 at 6:00am.

a stone is dropped from a cliff 200m high.if a second stone is thrown verticall upwards 1.50seconds after the first was released.strike the cliff base at the same instant as the first stone.with what velocity was the second stone thrown

  • mechanics - , Friday, February 1, 2013 at 5:27pm

    The First Stone.
    d1 = 0.5g*t^2 = 4.9*(1.5)^2 = 11 m. =
    Distance traveled by 1st stone after
    1,5 s.
    V^2 = Vo^2 + 2g*d.
    V^2 = 0 + 19.6*11 = 215.6
    V = 14.68 m/s = Velocity of 1st stone after 1.5 s.

    d1 = Vo*t + 0.5g*t^2 = 200-11.
    16.68t + 4.9t^2 = 189
    4.9t^2 + 16.68t - 189 = 0.
    Use Quadratic Formula.
    Tf = 4.74 s. = Fall time or time for each stone to reach Gnd.

    The 2nd Stone.
    d = Vo*t + 0.5g*t^2 = 200 m.
    Vo*4.74 + 4.9*(4.74)^2 = 200
    4.74Vo + 110.1 = 200
    4.74Vo = 200-110.1 = 89.9
    Vo = 19 m/s.

  • mechanics - , Friday, February 1, 2013 at 5:43pm

    Correction(TYPO).
    Change 16.68 to 14.68. Then Tf will be
    4.89 s.

    Change all 4.74s to 4.89.
    Vo = 89.9/4.89 = 18.4 m/s. = Initial velocity of 2nd stone.

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