Saturday
March 25, 2017

Post a New Question

Posted by on .

A baseball player releases a baseball at the height of 7 feet above the ground with an initial vertical velocity of 54 feet per second. How long will it take the ball to reach the ground?

  • algebra - ,

    Vf^2 = Vo^2 + 2gd,
    d = (Vf^2 - Vo^2) / 2g,
    d = 7 + (0 - (54)^2) / -64 = 52.56 Ft.
    = Distance above ground.

    Vf = Vo + gt,
    t(up) = (Vf - Vo) / g,
    t(up) = (0 - 54) / -32 = 1.69s.

    d = Vo*t + 0.5gt^2 = 52.56 Ft.
    0 + 0.5*32t = 52.56,
    16t = 52.56,
    t(down) = 3.29s.

    T = t(up) + t(down) = 1.69 + 3.29 = 4.98s = Time to reach ground.

  • algebra - ,

    Correction:
    0 + 0.5*32t^2 = 52.56,
    16t^2 = 52.56,
    t^2 = 3.29,
    t(down) = 1.81s.

    T = t(up) + t(down) = 1.69 + 1.81 = 3.50s. = Time to reach ground.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question