physics

You throw a baseball directly upward at time t = 0 at an initial speed of 13.3 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 9.80 m/s2.

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  1. v = Vi - g t
    at top v = 0
    0 = 13.3 - 9.81 t
    t = 1.36 seconds to top
    h = Hi + Vi t - 4.9 t^2
    h = 0 + 13.3 * 1.36 - 4.9 (1.36)^2
    h = 18.1 - 9.06
    h = 9.03 meters

    when is it at 4.52 meters?
    4.52 = 13.3 t - 4.9 t^2
    t^2 - 2.71 t + .922 = 0

    t = [ 2.71 +/- sqrt ( 7.34-3.69) ]/2
    t = [ 2.71 +/- 1.91 ]/2
    = .4 seconds on the way up
    and
    2.31 seconds on the way dwon

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