Answer:
Part (i) the initial acceleration of the rocket is 6.98 m/s²
Part(ii) the floor pushes on the power supply at 120m altitude by a force of 31.68 N
Explanation:
Part (i) the initial acceleration of the rocket.
For the rocket to accelerate, the force applied to it must overcome gravitational force due to its own weight.

Part(ii) how hard the floor pushes on the power supply at 120 m altitude
At 120 m height, the acceleration of the rocket is 6.98 m/s², which is the same as the power supply.
given force on power supply;
F = 18.5 N
Applying Newton's second law of motion, the mass of the power supply = 18.5/9.8
= 1.888 kg
The force on power supply at this altitude = m(a+g)
= 1.888(6.98 +9.8)
= 1.888(16.78)
= 31.68 N
Therefore, the floor pushes on the power supply at 120 m altitude by a force of 31.68 N