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 of81.0m/sat ground level. The engines then fire, and the rocket accelerates upward at4.20m/s2until it reaches an altitude of1160m. 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.)

Explanation / Answer

1st phase:- =>By v = u + at =>v = 79.8 + 4.10t1 -----------(i) By v^2 = u^2 + 2as =>(79.8 + 4.10t1)^2 = (79.8)^2 + 2 x 4.10 x 1070 =>(79.8)^2 + 16.81t1^2 + 654.36t1 = (79.8)^2 + 8774 =>16.8t1^2 + 654.36t1 - 8119.64 = 0 =>t1 = [-654.36 +/- v{(654.36)^2 - 4 x 16.8 x (-8119.64)}]/[2 x 16.8] =>t1 = [-654.36 +/- 986.83]/33.6 =>t1 = 9.89 sec Thus by (i) :- =>v = 129.37 m/s & By s = ut + 1/2at^2 =>h1 = 79.8 x 9.89 + 1/2 x 4.10 x (9.89)^2 =>h1 = 989.74 m 2nd Phase:- By v = u -gt =>0 = 129.37 - 9.8 x t2 =>t2 = 12.28 sec & By v^2 = u^2 - 2gh =>0 = (129.37)^2 - 2 x 9.8 x h2 =>h2 = 853.91 m 3rd phase:- By s = ut + 1/2gt^2 =>h1+h2 = 0 + 1/2 x 9.8 x t3^2 =>989.74+853.91 = 4.9t3^2 =>t3 = v376.25 =>t3 = 19.40 sec & By v = u + gt =>v = 0 + 9.8 x 19.40 =>v = 190.09 m/s (a) t = t1 + t2 + t3 = 41.57 sec (b)H = h1 + h2 = 1843.65 m (c) v = 190.09 m/s or D = AT^2 /2,,, D = 3.9 x T^2 /2 , so for T= 10 seconds, D = 195 meters + VT (790) = 790+195 = 985 meters V up = 79 + 39 = 118 m / s , time to stop = 10 + 12 = 22 s, D = 9.8x12^/2 = 710 + 970 = 1680 m Max Altitude = 1680 meters For down 1680 = 9.8 x T^2 / 2 = 18.51 seconds Total time = 10 + 12.69 + 18.51 = 41.2 seconds Velocity at impact = AT = 18.51 x 9.8 = 181.5 m /s

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