Answer:
Approximately [tex]486\; {\rm m}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
The rocket started from rest, so the initial velocity would be [tex]u_{0} = 0\; {\rm m\cdot s^{-1}}[/tex]. With an acceleration of [tex]a_{0} = 20.0\; {\rm m\cdot s^{-2}}[/tex], the displacement [tex]x_{0}[/tex] of the rocket at [tex]t_{0} = 4.00\; {\rm s}[/tex] will be:
[tex]\begin{aligned} x_{0}&= \frac{1}{2}\, a\, t^{2} \\ &= \frac{1}{2} \times 20.0 \times (4.00)^{2} \; {\rm m} \\ &= 160\; {\rm m}\end{aligned}[/tex].
In other words, the rocket would have reached a height of [tex]160\; {\rm m}[/tex] when the motors turns off.
The velocity of the rocket at that point will be:
[tex]\begin{aligned}v_{0} &= u_{0} + a_{0}\, t_{0} \\ &= 0\; {\rm m\cdot s^{-1}} + (20.0 \times 4.00)\; {\rm m\cdot s^{-1} \\ &= 80.0\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Since air resistance on the rocket is negligible, the acceleration of the rocket after motor power-off will be [tex]a_{1} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}[/tex] ([tex]g[/tex] is the strength of the gravitational field.)
The rocket started with a velocity of [tex]{u}_{1} = v_{0} = 80.0\; {\rm m\cdot s^{-1}}[/tex] when the motor powers off. When the rocket reaches maximum height, the velocity of the rocket will be [tex]v_{1} = 0\; {\rm m\cdot s^{-1}}[/tex]. Apply the SUVAT equation [tex]{v_{1}}^{2} - {u_{1}}^{2} = 2\, a_{1}\, x_{1}[/tex] to find the additional elevation [tex]x_{1}[/tex] that the rocket will gain after the motor powers off:
[tex]\begin{aligned} x_{1} &= \frac{{v_{1}}^{2} - {u_{1}}^{2}}{2\, a_{1}} \\ &= \frac{(0)^{2} - (80)^{2}}{2 \times (-9.81)} \\ &\approx 326\; {\rm m}\end{aligned}[/tex].
Combine [tex]x_{0}[/tex] (elevation at motor power-off) and [tex]x_{1}[/tex] (elevation height gained after motor power-off) to find the overall maximum elevation of the rocket:
[tex]x_{1} + x_{2} = (160 + 326)\; {\rm m} = 486\; {\rm m}[/tex].