# A rocket is fired from a plane flying horizontally at 9000 ft the height of the rocket above the plane is given by h=560t-16t^2 where t is the time when is the rocket more than 4000 ft above or below the plane

1
by sweetysiri92

2014-08-19T20:28:48+05:30

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H = 560 t - 16 t²                  we assume this equation gives height in feet.

Find the points of  time when H = 4000 ft above the plane

4000 = 560t - 16 t²      t² - 35 t + 250 = 0
Δ= 225        t = [35 +- 15]/2  = 25 sec or 10sec

find points of time when H = -4000 ft  = 560 t - 16 t²
t² -35t - 250 = 0          t = 41.08 sec

So the rocket crosses the height of 4000ft in 10seconds then reaches maximum top height at 17.5 sec and falls to 4000ft at 25 seconds, above the height of plane.
It then falls to the height of plane and next below plane at 4000ft at 41.08 sec

is it not a straight forward question? what is the difficulty
thanks & welcome
u r welcom & thanx