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 4.10 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?

(b) What is its maximum altitude?

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

Explanation / Answer

Apply kinematic equation between the point 0 to 1

vf^2 - vi^2 = 2as

vf^2 -(80.2)^2 = 2(4.10)(1040)

vf =122.31 m/s

vf= vi + at

122.31 = 80.2 + 4 .10 t1

t1 = 10.27 s

between the point 1 and 2

0-122.31 m/s = -9.8 t2

t2 = 12.48 s

the maximum altitude is

0-(122.31)^2 = 2 (-9.8) ( yf- yi)

yf- yi =763.25 m

from point 2 to 3

vf^2 -0 = 2 ( -9.8) (-1803.25)

vf =-187.99 m/s

vf= at

-187.99 m/s = - 9.8 t3

t3 = 19.18 s

(a)

Total time = t1 + t2 + t3 = 41.83 s

(b)

maximum alitude yf- yi = 1.8 km

(c)

vfinal = -187.99 m/s

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