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A ball is thrown upward from the top of a 24.4 m tall building. The ball's initial speed is 12 m/s. At the same instant, a person is running on the ground at a distance of 31.4 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?

  • physic -

    Tr = (Vf-Vo)/g = (0-12) / -9.8 = 1.22 s=
    Rise time or time to reach max .ht.

    hmax = ho + (Vf^2-Vo^2)/2g.
    hmax = 24.4 + (0-144) / -19.6 = 31.75 m=
    ht. above gnd.

    h = Vo*t + 0.5g*t^2 = 31.75 m.
    0 + 4.9t^2 = 31.75
    t^2 = 6.48
    Tf = 2.55 s. = Time to fall to gnd.

    D = r*(Tr+Tf).
    r = D/(Tr+Tf)
    r=31.4 / (1.22+2.55) = 8.33 m/s.

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