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 1100 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

v^2=u^2 +2*a*s ; v(at 1100m)=121.87 it moves up until final velocity becomes zero.using the same equation.

0=121.87^2 -2*(9.8)*s ;s=757.78m (total distance=1100+757.78)=1857.78m above ground

using the same formula (velocity will become zero for a moment as g will be downwards)

v^2=0 +2*9.8*1857.78 =190.82m/s(downwards)

time above the ground

v = u+at;t =10.94s now time until v=0 which is 12.43sec{0=121.87 - 9.8 t}

downward journey v(190.82)=0+9.8 t

19.47s

total(10.94+12.43+19.47)=42.85s

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