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A ball is thrown into the air with an upward velocity of 28 ft/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?
How many seconds would it take to descend 2 feet above the ground
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GPT-5 mini
AI
h(t) = −16t^2 + 28t + 6
1) Time to reach maximum (vertex):
t = −b/(2a) = −28/(2·−16) = 28/32 = 7/8 = 0.875 s.
2) Maximum height:
h(7/8) = −16(7/8)^2 + 28(7/8) + 6 = 18.25 ft (exactly 73/4 ft).
3) When h = 2 ft:
Solve −16t^2 + 28t + 6 = 2 → 4t^2 − 7t − 1 = 0
t = [7 ± √65]/8. Numerically: t ≈ −0.133 s (discard) and t ≈ 1.883 s.
So the ball reaches 2 ft above the ground at about 1.88 seconds after launch. (From the peak it takes ≈ 1.883 − 0.875 = 1.008 s to descend to 2 ft.)
1) Time to reach maximum (vertex):
t = −b/(2a) = −28/(2·−16) = 28/32 = 7/8 = 0.875 s.
2) Maximum height:
h(7/8) = −16(7/8)^2 + 28(7/8) + 6 = 18.25 ft (exactly 73/4 ft).
3) When h = 2 ft:
Solve −16t^2 + 28t + 6 = 2 → 4t^2 − 7t − 1 = 0
t = [7 ± √65]/8. Numerically: t ≈ −0.133 s (discard) and t ≈ 1.883 s.
So the ball reaches 2 ft above the ground at about 1.88 seconds after launch. (From the peak it takes ≈ 1.883 − 0.875 = 1.008 s to descend to 2 ft.)
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