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 80.6 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.90 m/s2 until it reaches an altitude of 1110 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

Given,

Initial velocity, u = 80.6 m/s

Acceleration, a = 3.90 m/s2

Distance, s = 1110 m

final speed = v2 = 80.62 + 2 x 3.9 x 1110

=> v= 123.10 m/s

Applying Newton's 2nd equation of motion,

1110 = (80.6 t) + 0.5 x 3.9 t2

t = 10.90 s

for free fall, t = 25.12 s ( 123.10 x 2= 9.8t)

for free fall downward = 7.04 s (1110 = 123.10 t+ 0.5 x 9.8 x t2)

total time = 43.06 s

b) 123.102 = 2 x 9.8 x S

S = 773.14 m

total height above ground = 1110 + 773.14 = 1883.14 m = 1.88 km

c) velocity just before hitting = 192.09 m/s ( 123.10 + 9.8 x 7.04)

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