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Physics

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A ball is thrown upward from the top of a 50.0 m tall building. The ball's initial speed is 12.0 m/s. At the same instant, a person is running on the ground at a distance of 44.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

  • Physics - ,

    Tr = (Vf-Vo)/g,
    Tr = (0-12) / -9.8 = 1.22 s. = Time to
    rise to max ht.

    hmax = ho + Vo*t + 0.5g*t^2,
    hmax=50 + 12*1.22 - 4.9(1.22)^2=57.35 m
    = ht. above gnd.

    hmax = Vo*t + 0.5g*t^2 = 57.35 m.
    0 + 4.9t^2 = 57.35,
    t^2 = 11.70,
    Tf = 3.42 s. = Time to fall to gnd.

    d = V*Tf,
    V = d/Tf = 44 / 3.42 = 12.9 m/s.=Speed
    of the person.

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