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A ball is dropped from rest from a height of 20.0m. One second later a second ball is thrown vertically downwards. If the two balls arrive on the ground at the same time, what must have been the initial velocity of the second ball?

  • Physics -

    Use the equations:
    S=vi*t + (1/2)gt^2

    For 1st ball:
    20=0+(1/2)gt^2 => t=sqrt(40/g)

    For 2nd ball:
    20 = vi*(t-1) + (1/2)g(t-1)^2
    vi = [20-(1/2)g(t-1)^2]/(t-1)
    =14.62 m/s downwards

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