A rocket is fired vertically upward from a well. A catapult gives it an initial

Question

A rocket is fired vertically upward from a well. A catapult gives it an initial velocity of 81.5 m/s at ground level. Its engines then fire and give it an upward acceleration of 4.30 m/s2 until it reaches an altitude of 1300 m. At that point, its engines fail and the rocket goes into free-fall, with an acceleration of -9.80 m/s2. How long is the rocket in motion above the ground? I got the the time for the first accelerated part of the flight to be 12.093 seconds, but i am stuck on the free-fall section of the flight. And once i get that how would i calculate the time for the rest of the flight?

Explanation / Answer

Hi, In first case when the engine is working, the time taken to reach 1300m can be found as follows. we know s = ut + (1/2)a(t)^2.       => 1300 = (81.5 * t) + (0.5 * 4.3 * t^2)       solving this we get t = 12.093sec = t1 say. Now at this time the final velcoity of the rocket is v = u + at = 81.5 + (4.3 * 12.093)                                                                                           = 133.4999 m/s. So now we can consider the second part of the problem as after the engine failed, which is now similar to a body thrown upwards from a height of 1300m with a velocity of 133.4999 m/s. The rocket will travel upwards for some time and come to rest momentarily at a distance say 'x' from this height given by       v^2 - u^2 = 2 * a * s    => s = (133.4999)^2 / (2 * 9.8) = 909.297m (since the acceleration acting on it now is only the accleration due to gravity). Hence the total distance travelled upwards = 1300 + 909.297 = 2209.297m The time taken to travel this distance = t2 is given from the formula       v = u + at => 0 = 133.4999 + (-9.8)*t2 (since the acceleration is downwards and rocket is moving downwards).     => t2 = 133.4999 / 9.8 = 13.623sec. Time taken for the downward fall from this height of 2209.297 to the ground t3 is given by equation s = ut + (1/2at^2) where in this case u = 0 since it starts from rest at the top and a = 9.8m/s. => 2209.297 = 0 + (0.5 * 9.8 * t3^2) = 4.9 * t3^2 => t3 = 21.234sec. Hence the total time the rocket in air = t = t1 + t2 + t3 = 12.093 + 13.623 + 21.234                                                               = 46.95sec Hope this helps you.               Hope this helps you.

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