I need c A catapult launches a test rocket vertically upward from a well, giving

Question

I need c

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.0 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.80 m/s^2 until it reaches an altitude of 990 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of -9.80 m/s^2. (You will need to consider the motion while the engine is operating and the free-fall motion separately For what time interval is the rocket in motion above the ground What is its maximum altitude? What is its velocity just before it hits the ground?

Explanation / Answer

a)

consider the motion when the engines were operating

Vo = initial velocity = 79 m/s

a = acceleration = 3.80

d = displacement upward = 990 m

t = time taken

using the equation

d = Vo t + (0.5) a t2

990 = 79 t + (0.5) (3.8) t2

t = 10.1 sec

velocity gained is given as

Vf = Vo + at = 79 + 3.8 (10.1) = 117.4 m/s

Consider the motion of rocket after engines fail until it reach the ground

Vo = initial velocity = 117.4 m/s

a = - 9.8

Y = displacement = - 990

t' = time taken while engine is not working

using the equation

Y = Vo t' + (0.5) a t'2

- 990 = 117.4 t' + (0.5) (- 9.8) t'2

t' = 30.6 sec

Total time = T = t + t' = 10.1 + 30.6 = 40.7 sec

c)

consider the motion after engine fails

Vf = Vo + at'

Vf = 117.4 + (-9.8) (30.6) = - 182.5 m/s

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