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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? |
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Answer» 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 Learn More: A body is dropped from a height h at t = 0, with zero INITIAL velocity. Its ... A BULLET is shot straight upward with an initial speed of 925 ft/s. The ... |
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