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Question
A model rocket is launched from the ground. The height h reached by the rocket after t seconds from lift off is given by h(t) = −5t2 +100t, 0 ≤ t ≤ 20. At what time the rocket is 495 feet above the ground?
Solution
H(t) = – 5t2 + 100t
At t = 0, h(0) = 0
At t = 1, h(1) = – 5 + 100 = 95
At t = 2, h(2) = – 20 + 200 = 180
At t =3, h(3) = – 45 + 300 = 255
At t = 4, h(4) = – 80 + 400 = 320
At t = 5, h(5) = – 125 + 500 = 375
At t = 6, h(6) = – 180 + 600 = 420
At t = 7, h(7) = – 245 + 700 = 455
At t = 8, h(8) = – 320 + 800 = 480
At t = 9, h(9) = -405 + 900 = 495
So, at 9 secs, the rocket is 495 feet above the ground.
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