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 80.0 m/s at ground level. The engine then fires and the rocket accelerates upward at 4.00 m/s^2 until it reaches an altitude of 1000m. At this point, it's engines fail and the rocket goes into free fall, with an acceleration of -9.8m/s^2.

(a) For what time interval is the rocket above the ground?

(b) What is the maximum altitude?

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

(You will need to consider the motion while the engine is operating and the free-fall motion separately).

Explanation / Answer

1000=80*t+2*t^2

500=40t+t^2

t^2+50t-10t-500=0

t1=10s

Then the engine fails it falls freely

uin=80+4*t1=80+40=120

-1000=120t-9.8t^2

t2= 17.93s

So the total time taken = t1+t2= 27.93s

b) Max altitude= 1000+120*120/(2*9.8)=1734.69 m

(c) v=sqrt(2*9.8*1734.69)=184.391 m/s

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