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A ball is thrown into the air with an upward velocity of 2828 ft/s. Its height h in feet after t seconds is given by the function hequals=negative 16 t squared plus 28 t plus 7−16t2+28t+7. 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 33 feet above the ground?
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The height function is h(t) = −16t^2 + 28t + 7.
1. Time of maximum height: use the vertex t = −b/(2a) with a = −16, b = 28:
t = 28/(32) = 7/8 = 0.875 s.
2. Maximum height: h(7/8) = −16(7/8)^2 + 28(7/8) + 7 = 77/4 = 19.25 ft.
3. Time(s) to descend to 33 ft: solve −16t^2 + 28t + 7 = 33 ⇒ 16t^2 − 28t + 26 = 0.
Discriminant = (−28)^2 − 4·16·26 = 784 − 1664 = −880 < 0, so there are no real solutions.
Thus the ball never reaches 33 ft (its maximum is only 19.25 ft).
If you meant a different initial velocity or a different height, tell me and I’ll recompute.
1. Time of maximum height: use the vertex t = −b/(2a) with a = −16, b = 28:
t = 28/(32) = 7/8 = 0.875 s.
2. Maximum height: h(7/8) = −16(7/8)^2 + 28(7/8) + 7 = 77/4 = 19.25 ft.
3. Time(s) to descend to 33 ft: solve −16t^2 + 28t + 7 = 33 ⇒ 16t^2 − 28t + 26 = 0.
Discriminant = (−28)^2 − 4·16·26 = 784 − 1664 = −880 < 0, so there are no real solutions.
Thus the ball never reaches 33 ft (its maximum is only 19.25 ft).
If you meant a different initial velocity or a different height, tell me and I’ll recompute.
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