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26 January, 08:32

If a ball is thrown directly upward with a velocity of 50ft/s, it's height (in feet) after t seconds is given by y=50t-16t^2 what is the maximum height attained by the ball

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  1. 26 January, 09:03
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    39.0625 ft

    Step-by-step explanation:

    Since the function is a quadratic representing height, and the coefficient of the t² is negative, the vertex of the parabola will be the maximum height achieved by the ball.

    The general form for a quadratic equation is ax² + bx + c, here a is - 16, and b is 50

    To find the x coordinate of the vertex, use x = - b / (2a)

    We have x = - 50/[2 (-16) ]

    x = - 50/-32

    x = 25/16

    Now plug that into the equation to find the y value, which will be the height ...

    y = 50 (25/16) - 16 (25/16) ²

    y = 1250/16 - 16 (625/256)

    y = 1250/16 - 625/16

    y = 625/16

    y = 39.0625
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