Answer
a) $t = 17.5 \space seconds$ or $t = 10 \space seconds$
b) The total time that passes before the rocket reaches the ground again is 17.5 + 10 = 27.5 seconds.
Work Step by Step
a) For $h = -16t^2 + 440t$,
When h = 2800 ft,
$2800 = -16t^2 + 440t$
$16t^2 - 440t + 2800 = 0$
$2t^2 - 55t + 350 = 0$
By quadratic formula,
$t = \frac{-(-55) \pm \sqrt{(-55)^2 - (4)(2)(350)}}{(2)(2)}$
$t = \frac{55 \pm 15}{4}$
$t = 17.5 \space seconds$ or $t = 10 \space seconds$
There are 2 answers because the rocket first reached 2800 feet at 10 seconds and then it continues to rise to its maximum whence it will fall back to 2800 feet again at 17.5 seconds.
b) The rocket will take 10 seconds to fall back to the ground from 2800 feet, assuming no resistance.
Hence, the total time that passes before the rocket reaches the ground again is 17.5 + 10 = 27.5 seconds.
