To physics keeners: Please show work below?

Question

Thanks in advance for your help!

A 500 g model rocket is resting horizontally at the top edge of a 40 m high wall when it is accidentally bumped. The bump pushes it off the edge with a horizontal speed of 0.5 m/s and at the same time causes the engine to ignite. When the engine fires, it exerts a constant 20 N horizontal thrust away from the wall.

Part A: How far from the base of the wall does the rocket land?
Part B: Describe the trajectory of the rocket while it travels the ground.

Answer 1

Part A.

First: Use m=0.5 kg, and F=20 N, substitute in equation F=ma, to find a = 40 m/s^2 . This is the horizontal acceleration.

Secondly: Use distance, s = 40 m, initial velocity,u=o and acceleration, a = 10 m/s^2 (it is assumed that this is the value of gravity), substitute in s=ut+1/2at^2 to find time, t. This equals 2.83 s. This is the time it takes to reach the ground. Note: you don't have to solve a quadratic equation to find t since the expression 'ut' equals 0.

Thirdly, use initial velocity, u=0.5 m/s, a=40 m/s^2, and t=2.83 s, substitute in the equation s=ut+1/2at^2 to find how far the rocket lands from the wall. You should get 161.593 m. Do you also have to take significant figures into consideration? Then the answer is 160 m (correct to 2 sig.fig.)

You can handle Part B on your own now.