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A ball is thrown into the air with an upward velocity of 28ft/s. Its height h in feet after t seconds is given by the function hequals=negative 16 t squared plus 28 t plus 6−16t2+28t+6. What is the maximum height the ball reaches? How long does it take for the ball to reach the maximum height? How long does it take for the ball to descend to 2 feet above the ground?
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GPT-5 mini
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h(t) = −16t^2 + 28t + 6
- Time of maximum: t = −b/(2a) = −28/(2·(−16)) = 7/8 s = 0.875 s.
- Maximum height: h(7/8) = 73/4 ft = 18.25 ft.
Time when h = 2:
Solve −16t^2 + 28t + 6 = 2 → 4t^2 − 7t − 1 = 0
t = (7 ± √65)/8. The positive physical time is t = (7 + √65)/8 ≈ 1.883 s.
(There is a negative root ≈ −0.133 s which is not physically relevant after launch.)
- Time of maximum: t = −b/(2a) = −28/(2·(−16)) = 7/8 s = 0.875 s.
- Maximum height: h(7/8) = 73/4 ft = 18.25 ft.
Time when h = 2:
Solve −16t^2 + 28t + 6 = 2 → 4t^2 − 7t − 1 = 0
t = (7 ± √65)/8. The positive physical time is t = (7 + √65)/8 ≈ 1.883 s.
(There is a negative root ≈ −0.133 s which is not physically relevant after launch.)
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