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)=-5t square+100t,0less than or equal to t less than or equal to 20. At what time the rocket is 495 feet above the ground
Answers
Given : A model rocket is launched from the ground. The height h reached by the rocket after t seconds
h(t)= -5t²+100t
To Find : At what time the rocket is 495 feet above the ground?
Solution:
h(t)=-5t²+100t
Let say at time = x rocket is 495 feet above the ground
=> h(x) = 495
h(x) = -5x² + 100x
-5x² + 100x = 495
=> x² - 20x = -99
=> x² - 20x + 99 =0
=> x² - 9x - 11x + 99 = 0
=> x(x - 9) - 11(x - 9) = 0
=> (x - 9) ( x - 11) = 0
=> x = 9 , x = 11
rocket is 495 feet above the ground at 9 sec & 11 sec
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Answer:
Solution:
h(t) = -5t² + 1oot
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.