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 80.2 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 1040 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?
_______ _______
Your response differs from the correct answer by more than 10%. Double check your calculations. s

(b) What is its maximum altitude?
_______________________
Your response differs from the correct answer by more than 10%. Double check your calculations. km

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

Explanation / Answer

(a) time taken to reach the 1040 m.

d = v0t + a t^2 /2

1040 = 80.2t + 3.90 t^2 / 2

1.95 t^2 + 80.2t - 1040 = 0

t1 = 10.36 sec

after that:

vertical displacement as it reaches the ground.

y = - 1040 m

v0 = 80.2 + (10.36 x 3.90) = 120.6 m/s

y = v0t - g t^2 /2

- 1040 = 120.6 t - 9.8 t^2 / 2

4.9 t^2 - 120.6t - 1040 = 0

t2 = 31.38 sec

t = t1 + t2 = 41.7 sec .......Ans

(B) maximum altitude will be when velocity is zero.

vf^2 - vi^2 = 2 a d

0^2 - 120.6^2 = 2 x -9.8 x h

h = 742 m

Hmax = 742 + 1040 = 1782 m ........Ans

(C) vf = v0 - g t

v = 120.6 - (9.8 x 31.38)

v = - 187 m/s

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