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.)

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 :)

To find the time it takes for the rocket to reach an altitude of 1180 m, we need to consider two separate phases of its motion: the phase when the engine is operating, and the phase when it goes into free fall.

1. Motion while the engine is operating:
We can use the kinematic equation to determine the time it takes for the rocket to reach an altitude of 1180 m while the engine is operating. The equation we can use is:
Δy = v_i*t + (1/2)*a*t^2

In this equation, Δy is the displacement (1180 m), v_i is the initial velocity (80.2 m/s), a is the acceleration (3.80 m/s²), and t is the time.

Rearranging the equation, we have:
Δy - (1/2)*a*t^2 = v_i*t

Substituting the given values, we get:
1180 - (1/2)*3.8*t^2 = 80.2*t

Rearranging and simplifying:
1.9*t^2 - 80.2*t + 1180 = 0

This is a quadratic equation, so we can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values, we get:
t = (-(-80.2) ± √((-80.2)^2 - 4*(1.9)*(1180))) / (2*(1.9))

Solving the quadratic equation, we find two roots: t1 ≈ 13.55s and t2 ≈ 46.12s.
Since we are interested in the positive value, the time it takes for the rocket to reach an altitude of 1180 m while the engine is operating is approximately 13.55 seconds.

2. Free-fall motion:
After the engines fail, the rocket goes into free fall. In this phase, the acceleration is -9.80 m/s² (negative because it is directed downwards).

We can use the kinematic equation again to find the time it takes for the rocket to reach the highest point while in free fall. Since we want to find the time it takes for the rocket to fall back to the ground, we can double the time.

Δy = v_i*t + (1/2)*a*t^2

Rearranging the equation, we have:
Δy = (1/2)*a*t^2

Substituting the given values, we get:
1180 = (1/2)*(-9.8)*t^2

Simplifying:
-4.9*t^2 = 1180

Dividing both sides by -4.9:
t^2 ≈ -240.82

Since time cannot be negative, we discard this solution. This means the rocket never falls back down after reaching an altitude of 1180 m.

Therefore, the total time it takes for the rocket to reach an altitude of 1180 m and come back down to the ground is approximately 13.55 seconds.