Acatapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level. the engines then fire, and the rocket accelerates upward at 3.90 m/s2 until it reaches an altitude of 1040 m. at that point its engines fail, and the rocket goes into free fall, with an acceleration of −9.80 m/s2. (you will need to consider the motion while the engine is operating and the free-fall motion separately.) (a) for what time interval is the rocket in motion above the ground?

solution:

y = v0t + ½at²

1150 = 79t + ½3.9t²

0 = 3.9t² + 158t - 2300

from quadratic equations and eliminating the negative answer

t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)

t = 11.37 s to engine cut-off

the velocity at that time is

v = v0 + at

v = 79 + 3.9(11.37)

v = 123.3 m/s

it rises for an additional time

v = gt

t = v/g

t = 123.3 / 9.8

t = 12.59 s

gaining more altitude

y = ½vt

y = 123.3(12.59) /2

y = 776 m

for a peak height of

y = 776 + 1150

talk back to him about the sport and show him your expertise or leave the room

i think its c if i'm wrong