A rocket designed to place small payloads into orbit is carried to an altitude o

Question

A rocket designed to place small payloads into orbit is carried to an altitude of 12000 m by an aircraft. The aircraft flies straight and level at 850 km/h and drops the rocket. The aircraft then maintains its speed and altitude. The rocket drops, and ftera time t ( which you will calculate) its rocket engine ignites. Once it ignites the rocket accelerates at a=3.00g at an angle 30 degrees above the horizontal under the combined effects of its engine and gravity. For safety, the rocket must be at least 1000 m ahead of the aircraft when it passes back through the aircraft's altitude. Ignore air resistance. a) Draw an x vs t graph of both rocket and aircraft, with as much detail as you can, include an indication of when the rocket is dropped, when the engine ignites, and when the rocket climbs through the aircraft's altitude. b) Draw a y vs t graph of both rocket and aircraft with as much detail as you can. Include an indication of when the rocket is dropped, when the engine ignites, and when the rocket climbs through the aircraft's altitude c) determine the minimum time that the rocket must fall in free-fall before the engine ignites.

Explanation / Answer

Dropping units for ease of typing; all units SI (kg/m/s) Vertically: airliner: h(t) = 0 (let's call this the reference altitude) Easy enough. But the rocket is another story. It will drop for some time t, after which it will have attained a negative velocity of Vo = -gt and a negative displacement of Ho = -½gt². Then the motor will turn on, and the altitude (relative to 12000m) can be determined by H(t) = Ho + Vo•t + ½•3•9.8•sin30•t² = Ho + Vo•t + 7.35t² where t is the duration of motor burn and Ho and Vo, both negative, are calculated as shown above. Horizontally: First, let's convert the velocity to SI: 850km/hr • 1000m/km • 1hr/3600s = 236 m/s airliner: s(t) = 236t (at end of rocket free-fall), and s(t) = 236t + 236t (during burn) rocket: Kind of the same deal: during free-fall, S(t) = 236t (at end of free-fall), and S(t) = 236t + 236t + ½•3•9.8•cos30•t² = 236t + 236t + 12.7t² (during burn) Whew. OK, what do we know? We need S(t) = s(t) + 1000m, so let's solve for the equality: 236t + 236t + 12.7t² = 236t + 236t + 1000 first two terms on each side cancel, leaving 12.7t² = 1000 t = 8.9s What is that? That's time time after rocket ignition that it takes to get 1000m ahead of the airliner. Where does that get us? Since we're interested in where the trajectories intersect, let's assume that after 8.9s the rocket has reached reference height 0: H(8.9) = 0 = Ho + Vo•8.9 + 7.35•8.9² = Ho + 8.9Vo + 579 Substitute in for Ho and Vo: 0 = -½•9.8•t² - 8.9•9.8t + 579 = 4.9t² + 87.2t - 579 quadratic t = (-b ± sqrt(b² - 4ac) ) / 2a = (-87.2 ± sqrt(87.2² + 4•4.9•579) ) / (2•4.9) t = (-87.2 ± 137.7) / 9.8 ? discard -ve root t = 5.15 s ? answer Wow. Let's do a little checking. Ho = -½g5.5² = -½•9.8•5.5² = -148 m Vo = -gt = -9.8•5.15 = -50 m/s H(8.9) = Ho + Vo•8.9 + 7.35•8.9² = -148 + 8.9•(-50) + 579 = -14 m is close enough for this man's army.

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