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 3.90 m/s2 until it reaches an altitude of 1140 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? (Iv tried this one over and over again and cant figure it out)

Explanation / Answer

1)INITIAL VELOCITY=80.8m/s time to reach 1140 is 1140=80.8t+0.5*3.9*t^2 t=11.12sec v @ t=11.12 is=124.168 hmax=1140+(124.168)^2]/2g=1910 time to reach hmax=11.12+12.67 time for down travel=19.74sec total=43.533sec plss rate

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