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22 March, 15:35

A ball is thrown into the air with an upward velocity of 20 feet per second it's height h in feet after t seconds is given by the function h (t) = 16t^2+20t+2. How long does it take the ball to reach its maximum height? What is the ball's maximum height?

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  1. 22 March, 18:22
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    The extreme of the quadratic ax² + bx + c is found at x = - b / (2a). Your function has

    a = 16

    b = 20

    so it will have an extreme value at t = - 20 / (2*16) = - 20/32 = - 5/8.

    We assume you intend h (t) = - 16t² + 20t + 2 (with a leading minus sign), in which case the extreme value (maximum height) occurs at t = 5/8. The maximum height is

    h (5/8) = (-16*5/8 + 20) (5/8) + 2 = 50/8 + 2 = 33/4 = 8.25

    The ball reaches its maximum height after 5/8 = 0.625 seconds.

    The ball's maximum height is 8 1/4 = 8.25 feet.
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