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A ball is thrown upward from the ground with an initial speed of 23.2 m/s; at the same instant, another ball is dropped from a building 20 m high. After how long will the balls be at the same height?

  • Physics -


    Let the first ball have height x, velocity u.
    x(0) = 0[m]
    u(0) = 23.2[m/s]
    x(t) = x(0) + u(0) t + (g/2) t^2
    .: x(t) = 23.2 t - 4.9 t^2

    Let the second have height y, velocity v.
    y(0) = 20[m]
    v(0) = 0[m/s]
    y(t) = y(0) + v(0) t + (g/2) t^2
    .: y(t) = 20 - 4.9 t^2

    Where: g = -9.8[m/s^2]

    Find: t such that x(t)=y(t)

  • Physics -

    I set the two position equations together and got 0.86 s as my time. Thanks!

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