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4 March, 20:14

From 4 feet above a swimming pool Susan throws a ball up word with a velocity of 32 ft./s. The height h (t) of the ball t seconds after Susan throws it is given by h (t) = -16t^2+32t+4. For >_0 find the maximum height reached by the ball at the time that the height is reached.

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  1. 4 March, 20:37
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    h (t) = 20 ft

    Step-by-step explanation:

    The equation for the trajectory is according to problem statement

    h (t) = - 16*t² + 32*t + 4

    Then velocity is dh/dt

    d (h) / dt = - 32*t + 32

    at h max d (h) / dt = 0

    - 32*t + 32 = 0

    t = 1 sec

    Therefore after 1 second the ball reachs the maximum height

    plugging this value in the equation of trajectory we get maximum height

    h (t) = - 16*t² + 16*t + 4

    h (t) = - 16 * (1) ² + 32*1 + 4

    h (t) = 20 ft from swimming pool level
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