Given the function h(t) where t is the time, to find the maximum height reached by the ball, we must find the maximum value of h(t) = -16t² + 46t + 6. Since h(t) is a quadratic function, to find its maximum value, we need to look for the coordinates of its vertex (x, y). To solve for the x-coordinate, we have
![x = \frac{-b}{2a}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%7B-b%7D%7B2a%7D%20)
![x = \frac{-46}{2(-16)} = \frac{23}{16}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%7B-46%7D%7B2%28-16%29%7D%20%3D%20%5Cfrac%7B23%7D%7B16%7D%20)
Thus, it takes t = 23/16 s for the ball to reach the maximum height. To find the maximum height, we need to substitute the value of t into h(t).
h = -16(23/16)² + 46(23/16) + 6
h = 625/16 ft
For the ball to reach ground level, then h must be equal to zero. Thus, we have
0 = -16t² + 46t + 6
0 = 4t² - 23t - 3
Recalling the quadratic formula, we have
![t = \frac{-b \pm \sqrt{b^{2}-4ac} }{2a}](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7B-b%20%5Cpm%20%5Csqrt%7Bb%5E%7B2%7D-4ac%7D%20%7D%7B2a%7D%20)
where a , b, and c are the coefficients of the quadratic equation. For our case, a = 4, b = -23, and c = -3. Thus, we have
![t = \frac{23 \pm \sqrt{23^{2}-4(4)(-23)} }{2(4)}](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7B23%20%5Cpm%20%5Csqrt%7B23%5E%7B2%7D-4%284%29%28-23%29%7D%20%7D%7B2%284%29%7D%20)
![t = \frac{23 \pm \sqrt{897}}{8}](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7B23%20%5Cpm%20%5Csqrt%7B897%7D%7D%7B8%7D%20)
Since time cannot be negative, we disregard the negative root and have t = (23 + √897)/8 s.
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Answer: </span><span>
Maximum time = 23/16 seconds</span><span>
Maximum height = 625/16 ft</span>
Time taken to reach the ground = (23 + √897)/8 seconds