Physics

You throw a baseball directly upward at time t=0 at an initial speed of 13.1 m/s. What is the maximum height the ball reaches above where it leaves your hand? Ignore air resistance and take g=9.80 m/s2.
maximum height:

At what times does the ball pass through half the maximum height?

earlier time at half maximum height:

later time at half maximum height:

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  1. h = -4.90 t^2 + 13.1 t

    t-max is on the axis of symmetry ... t = -b / 2a = -13.1 / (2 * -4.90)
    ... plug the time into the equation to find the max height

    plug half the max height into the equation to find the times (up and down)

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