Answer:
t = 16.5 s
Explanation:
First we apply first equation of motion to the accelerated motion of the rocket:
[tex]v_{f1} = v_{i1} + at_{1}[/tex]
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,
[tex]v_{f} = 0\ m/s + (30\ m/s^2)(4\ s)\\\\v_{f} = 120\ m/s[/tex]
Now, we analyze the motion rocket when engine turns off. So, the rocket is now in free fall motion. Applying 1st equation of motion:
[tex]v_{f2} = v_{i2} + g t_{2}[/tex]
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,
[tex]0\ m/s = 120\ m/s + (- 9.8\ m/s^2)(t_{2})\\\\t_{2} = \frac{120\ m/s}{9.8\ m/s^2}\\\\t_{2} = 12.25\ s[/tex]
So, the time taken from the firing position till the stopping position is:
[tex]t = t_{1} + t_{2}\\\\t = 4 s + 12.5 s[/tex]
t = 16.5 s
