Answer:
t = 16.5 s
Step-by-step explanation:
First we apply first equation of motion to the accelerated motion of the rocket:

where,
vf₁ = final speed of rocket during accelerated motion = ?
vi₁ = initial speed of rocket during accelerated motion = 0 m/s
a = acceleration of rocket during accelerated motion = 30 m/s²
t₁ = time taken during accelerated motion = 4 s
Therefore,

Now, we analyze the motion rocket when engine turns off. So, the rocket is now in free fall motion. Applying 1st equation of motion:

where,
vf₂ = final speed of rocket after engine is off = 0 m/s
vi₂ = initial speed of rocket after engine is off = Vf₁ = 120 m/s
g = acceleration of rocket after engine is off = - 9.8 m/s² (negative sign for upward motion)
t₂ = time taken after engine is off = ?
Therefore,

So, the time taken from the firing position till the stopping position is:

t = 16.5 s