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 4.10 m/s until it reaches an altitude of 910 m. At that point its engines fall, 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 )

Y = u t + 1/2 a t2  

910 = 79.2 X t + 0.5 X 4.1 X t2

2.05 t2 + 79.2 t -910 = 0

t = 9.26 sec

so we have v = u + a t

v = 79.2 + 4.1 X 9.26

v = 117.166 m/sec

and v = g t

t = v / g = 117.166 / 9.8

t = 11.955 sec

the gianing altitude is Y = 0.5 X v t

Y = 0.5 X 117.166 X 11.955

Y = 700.35 m

b )

so the total distance is dy = 700.35 + 910 = 1610.35 m

we have Y = 0.5 X g t2

then t2 = 2 X dy / g

t2 = 2 X 1610.35 / 9.8

t2 = 328.64 sec

t = 18.12 sec

then T = 18.12 + 11.955 + 9.26

T = 39.335 sec

c )

v = g T

v = 9.8 X 18.12

v = 177.576 m/sec

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