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16 October, 01:22

The equation h=-16t^2 + 32t + 9 gives the height of a ball, h, in feet above the ground, at t seconds after the ball is thrown upward. How many seconds after the ball is thrown will it reach maximum height? What is its maximum height?

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  1. 16 October, 04:37
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    H = - 16 t² + 32 t + 9

    The maximum height is at the vertex of the parabola:

    t = - b / (2 a) = - 32 : ( - 32) = 1 s

    The ball will reach maximum height in 1 second.

    h max = - 16 * 1² + 32 * 1 + 9 = - 16 + 32 + 9 = 25 m

    The maximum height is 25 m.
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