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 3.90 m/s2 until it reaches an altitude of 1060 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

Initial Speed u = 79.2m/s

Acceleration a =3.90 m/s^2

s = 1060 m

S = u*t + 0.5 at^2

1060 = 79.2 *t + 0.5*3.9 *t^2

1.95 t^2 + 79.2 t - 1060 = 0

Solving for t

t = 10.61 s

V = u + a*t

V = 79.2 + 3.9 * (10.61)^2

V = 518.23 m/s

At this point engine fails -

V = u -gt

0 = 518.23 - 9.8 * t

518.23 = 9.8 * t

t = 518.23/9.8 = 52.88s

Height which rocket rises with engine off -

s = 518.23 * 52.88 - 0.5 * 9.8 * (52.88)^2

s = 13702.16 m

Now , It starts falling Again.

Initial velocity = 0

Final Velocity = v

s = 13702.16 + 1060 = 14762 m

a = 9.8 m/s^2

s = ut + 0.5 at^2

14762 = 0.5 * 9.8 * t^2

t = 54.88 s

v = a*t

v = 9.8 * 54.88

v = 537.84 m/s

a)
Time interval rocket is in motion above the ground =10.61 s + 52.88s + 54.88 s
Time interval rocket is in motion above the ground = 118.37 s

b)
Maximum Altitude reached = 13702.16 + 1060
Maximum Altitude reached = 14762 m

c)
Velocity just before it hits the ground v = 537.84 m/s

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