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)=-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​

Answer 1

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 2

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.