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 4.20 m/s2 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/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

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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 1010 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

Answer

Part A)

The distance during the acceleration leads us to velocity when the engines fail

vf^2 = vo^2 + 2ad

vf^2 = (80.8)^2 + 2(4.1)(1010)

vf = 121.7 m/s

The time for this part...

vf = vo + at

121.7 = 80.8 + (4.1)(t)

t = 9.98 sec

The distance it will then travel...

vf^2 = vo^2 + 2ad

0 = (121.7)^2 + (2)(-9.8)(d)

d = 755.6 m

The time to travel to that height

vf = vo + at

0 = 121.7 + (-9.8)(t)

t = 12.4 sec

Then it will fall 1010 + 755.6 m = 1765.6 m

d = vot + .5at^2

1765.6 = 0 + (.5)(9.8)t

t = 19.0sec

Total time is 9.98 + 12.4 + 19.0 = 41.4 sec

Part B)

Solved in part A to be 1765.6 m

Part C)

vf^2 = vo^2 + 2ad

vf^2 = (0) + 2(9.8)(1765.6)

vf = 186 m/s

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