Answer:
Approximately [tex]72.9\; \rm m[/tex], assuming that the rocket had no propulsion onboard, and that air resistance on the rocket is negligible.
Explanation:
Initial velocity of this rocket: [tex]u = 27\; \rm m\cdot s^{-1}[/tex].
When the rocket is at its maximum height, the velocity of the rocket would be equal to [tex]0[/tex]. That is: [tex]v = 0\; \rm m \cdot s^{-1}[/tex].
The acceleration of the rocket (because of gravity) is constantly downwards, with a value of [tex]a = -g = -10\; \rm m \cdot s^{-2}[/tex].
Let [tex]x[/tex] denote the distance that the rocket travelled from the launch site to the place where it attained maximum height. The following equation would relate [tex]x \![/tex] to [tex]u[/tex], [tex]v[/tex], and [tex]a[/tex]:
[tex]\displaystyle x = \frac{v^2 - u^2}{2\, a}[/tex].
Apply this equation to find the value of [tex]x[/tex]:
[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{\left(0\; \rm m\cdot s^{-1}\right)^{2} - \left(27\; \rm m \cdot s^{-1}\right)^{2}}{2 \times 10\; \rm m \cdot s^{-2}} = 36.45\; \rm m\end{aligned}[/tex].
In other words, the maximum height that this rocket attained would be [tex]36.45\; \rm m[/tex].
Again, assume that the air resistance on this rocket is negligible. The rocket would return to the ground along the same path, and would cover a total distance of [tex]2\times 36.45\; \rm m = 72.9\; \rm m[/tex].
