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 80.8 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.80 m/s2 until it reaches an altitude of 930 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

time taken to reach 930m is 930=80.8t+0.5*3.8*t^2 we get t=9.422sec velocity of rocket at this point=80.8=3.4*9.422=112.83 the height it reaches from this point with 112.83m/sec is=(112.83^2/2g)=649.58m time taken to reach this point is 11.513sec total uptravel time=11.513+9.4222=20.935sec total downtravel time=sqrt(2h/g) h=649.58+930=1579.58 time downtravel=17.9544sec total time=38.889sec b)hmax=1579.58m c)velocity=sqrt(2g*hmax)=175.95m/s

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