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Two diamonds begin a free fall from rest from the same height 0.9 s apart. How long after the first object begins to fall will the two objects be 11 m apart?

  • physics -

    Let t=0 be the time the first one is dropped. The distance that it falls after that is
    y1 = (g/2)t^2 = 4.9 t^2
    The distance that the second one has fallen at time t(>0.9s) is
    y2 = (g/2)(t-0.9)^2 = 4.9(t-0.9)^2

    You want to solve for the time t when
    y1 - y2 = 4.9 [t^2 - (t-1)^2] = 11 m

    2t - 1 = 11/4.9 = 2.24 s
    t = 1.62 s

  • physics -

    Two diamonds begin a free fall from rest from the same height 1.2 s apart. How long after the first object begins to fall will the two objects be 20 m apart?

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