A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.2 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.80 m/s2 until it reaches an altitude of 1180 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.)

Answer 1

well during the acceleration up to 1180:
Vi = 80.2 +3.80 t
1180 = 80.2 t + (1/2)(3.80) t^2
solve for t and Vi
now the fall
v = Vi - 9.80 t (note where v = 0 is the top)
h = 1180 + Vi t - 4.9 t^2
You did not say what is asked but if you put in h = 0 you can solve for time falling and v at the ground (negative I trust :)