During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth's surface and is to reach a maximum height of 980 mm above the earth's surface. The rocket's engines give the rocket an upward acceleration so it moves with acceleration of 16.0 m/s2m/s2 during the time TT that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance. Assume that the acceleration due to gravity does not change with the height of the rocket.

   time tt = 40.76 s

Explanation:

In this exercise we can find the acceleration time, the speed with which the acceleration ends and the rise and flight time

For all these calculations we can use vertical launch kinematics

The speed for maximum height is zero

          v² = v₀² - 2 g y

          v₀ = √ 2gy

          v₀ = √ (2 9.8 980)

          v₀ = 138.59 m / s

The time it takes to climb after finishing the acceleration is

          v = v₀ - gt

          t₁ = v₀ / g

          t₁ = 138.59 / 9.8

          t₁ = 14.14 s

The acceleration time is

          v = v₀ + a t₂

   As part of the rest the initial speed is zero

          t₂ = v / a

          t₂ = 138.59 / 16

          t₂ = 8.66 s

The height raised at this time is

         y₂ = v₀ t + ½ a t²

         y₂ = 0 + ½ 16 8.66²

         y₂ = 600 m

Let's look for the descent time, the height of the rocket is

         y = y₂ + y₁

         y = 600 + 980

         y = 1580 m

         y = v₀t + ½ g t²

         v₀ = 0

         t₃ = √ 2y / g

         t₃ = √ (2 1580 / 9.8)

         t₃ = 17.96 s

Total flight time is

         tt = t₁ + t₂ + t₃

         tt = 14.14 + 8.66 + 17.96

         tt = 40.76 s

the answer would be   dignified - apex