Acatapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level. the engines then fire, and the rocket accelerates upward at 4.20 m/s2 until it reaches an altitude of 1180 m. at that point its engines fail, and the rocket goes into free fall, with an acceleration of −9.80 m/s2. (you will need to consider the motion while the engine is operating and the free-fall motion separately.) (a) for what time interval is the rocket in motion above the ground?

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

Time taken to reach 1180 m is 11.29 seconds

Time the rocket will keep going up after the engines shut off is 13.06 seconds.

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

Time taken by the rocket to fall from maximum height is 20.29 seconds

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s