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Calculus

A ball is thrown straight up from the top of a hill 30 feet high with an initial velocity of 72ft/sec. How high above level ground will the ball get? (Objects subjected to gravity adhere to s(t) = -16t^2 + v0t + so where s is the height of the object in feet v0 is the initial velocity and s0 is the initial height)

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  1. Your equation will be
    s(t) = -16t^2 + 72t + 30

    We need the vertex of this parabola, the t of the vertex is -b/2a
    or -72/-32 = 2.25 seconds
    s(2.25) = -16(2.25)^2 + 72(2.25) + 30
    = ..... m

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