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 79.2 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.80 m/s2 until it reaches an altitude of 1150 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? _____s

(b) What is its maximum altitude? _____ km

(c) What is its velocity just before it hits the ground?_____ m/s

Explanation / Answer

a)

During accelerated motion :

a = 3.80

Vo = 79.2 m/s

d = 1150 m

t1 = time during rocket active

using the equation

d = Vo t1 + (0.5) a t12

1150 = 79.2 t1 + (0.5) (3.80) t12

t1 = 11.4 sec

using the equation :

velocity at altitude 1150 = Vf = Vo + at = 79.2 + 3.80 (11.4) = 122.52 m/s

For free fall :

d = - 1150

Vo = 122.52 m/s

t2 = time of free fall

a = - 9.8 m/s2

using the equation

d = Vo t2 + (0.5) a t22

- 1150 = 122.52 t2 + (0.5) (- 9.) t22

t2 = 32.3 sec

total time = t = 11.4 + 32.3 = 43.7 sec

b)

h = height above the point at which engine fails

Vf2 = Vo2 + 2 a h

02 = 122.522 + 2 (-9.8) h

h = 765.88 m

Total height = 765.88 + 1150 = 1915.88

c)

Vf = Vo + at = 122.52 + (- 9.8) (32.3) = 194.02 m/s

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