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 80.8 m/s at ground level. The engines then fire, and the rocket accelerates upward at 4.10 m/s2 until it reaches an altitude of 1160 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?

(b) What is its maximum altitude?

(c) What is its velocity just before it hits the ground?

Explanation / Answer

When engine stops there will be some velocity, it will go upward till this velocity becomes zero then it comes to ground.

Velocity at that time, V2 - u2 = 2aS where S is displacement.

V = 126.65m/s.

Time to reach here will be 11.18 seconds.

After height reached to become velocity zero, in 818.37 m

Hence total altitude will be 1978.37m.

For time, 11.18 + time in which velocity becomes zero + time in which it comes down = 11.18 + 126.65/9.80 ( V = u + at) + 20.09 ( s = 1/2 at2) = 44.197 seconds.

Velocity, V = at = 196.88m/sec

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