Post a New Question

Physics

posted by on .

A ball is thrown upward from the top of a 24.6 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 31.2 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 - ,

    hmax = ho + (V^2-Vo^2)/2g.
    hmax = 24.6 + (0-144)/-19.6 = 31.95 m
    Above gnd.

    hmax = Vo*t + 0.5g*t^2 = 31.95 m.
    0 + 4.9t^2 = 31.95
    t^2 = 6.52
    Tf = 2.55 s. = Fall time.

    d = V*Tf = 31.2 m.
    V * 2.55 = 31.2
    V = 12.22 m/s.

  • Physics - ,

    Correction:
    Tr = (V-Vo)/g = (0-12)/-9.8 = 1.22 s. =
    Rise time.

    hmax and Tf are correct as shown in the
    previous analysis.

    d = V * (Tr+Tf) = 31.2 m.
    V * (1.22+2.55) = 31.2
    V = 8.28 m/s.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question