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A baseball player throws a baseball up at 24.3 m/s. How long must he wait before he can catch the baseball on its way down?

  • Physics -

    y = y0 + vy*t + 1/2*ay*t^2

    Where y is the distance in the vertical direction, y0 is the initial y position, vy is the speed in the y direction, ay is the acceleration in the y direction, t is time.

    Choose a coordinate system so that y0 equals zero at the baseball player's hand. y0=0; vy=24.3; ay = -9.8

    0 = 24.3*t - 1/2*9.8*t^2
    Dividing by t:

    0 = 24.3 - 1/2*9.8*t
    Solve for t --this is the time at which the ball reaches 0 again, or the same height from which it was released

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