Answer:
[tex]\large \boxed{\text{20.4 s}}[/tex]
Explanation:
I assume the rocket is going straight up. Then we need to deal only with the vertical direction.
Start with the equation
[tex]v = v_{0} + at[/tex]
At t = 0, v₀ = 100 m/s.
At time t, at the height of the trajectory, v = 0.
a = g = -9.807 m·s⁻²
[tex]\begin{array}{rcl}0 & = & 100 - 9.807t \\-100& = &-9.807t\\t & = & \dfrac{100}{9.807}\\\\& = & \text{10.20 s}\\\end{array}[/tex]
This is the time it takes to reach maximum height.
It will take the same time to reach the ground.
Thus,
Time of flight = 2t = 2 × 10.20 s = 20.4 s
[tex]\text{It will take $\large \boxed{\textbf{20.4 s}}$ for the rocket to return to the launch level.}[/tex]
