A catapult launches a test rocket vertically upward from a well, giving the rock

Question

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level. The engines then fire, and the rocket accelerates upward at 4.00 m/s2 until it reaches an altitude of 1 000 m. At that point, its engines fail and the rocket goes into free fall, with an acceleration of -9.80 m/s2. (a) For what time interval is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it hits the ground? (You will need to consider the motion while the engine is operating and the free-fall motion separately.)

Explanation / Answer

V^2 = U^2 + 2 .a . d

V^2 = 80^2 + 2 . 4 . 1000 = 14400

Final velocity ( at engine cutoff ) = 120 m/s

a) Time to go from 80 m/s to 120 m/s = 40 / 4 = 10 s

Extra Distance to top of flight = V^2 / 2 g = 120^2 / 2 . 9.8 = 734.69 m

b) Maximum altitude = 1734.69 m = 1.73 km.

Extra time needed to reach this height = 120 / 9.8 = 12.244 s

Time to fall 1.73 km = 2d / g = 2 . 1734.69 / 9.8 = 18.81s

Total time in motion = 18.81 + 12.244 + 10= 41.054 s

c) Speed after falling 1.73 km = 2 . g. d = 184.39m/s

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