What Is the Landing Distance of the Enmeshed Spy Plane and Rocket?

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The discussion centers on calculating the landing distance of a 575 kg unmanned spy plane and a 1280 kg rocket after they become enmeshed following an interception. The initial calculations utilized conservation of momentum and the Pythagorean theorem, yielding a velocity of 519.35 m/s. The time of descent was calculated to be 23.46 seconds, leading to an initial horizontal distance of 1.22 x 10^4 meters. However, the final distance from the point of impact was incorrectly calculated, prompting participants to seek clarification on the correct methodology and answer.

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In a military test, a 575 kg unmanned spy plane is traveling north at an altitude of 2700 m and a speed of 450 m/s. It is intercepted by a 1280 kg rocket traveling east at 725 m/s. If the rocket and spy plane become enmeshed in a tangled mess, how far from the point of impact do they hit the ground?

O.K. I take the plane is traveling northward parrallel to the ground. Conservation of momentum- and pythagorean theorum yields-
sqrt((1280 x 725)^2+(450 x 575)^2) = (1280 +575)V
Solving for V gives 519.35, and it is now a projectile motion. Solving for the time would give
-2700 = -4.905t^2
which gives time 23.46
then, I multiplied velocity x time, which gave 1.22 x 10^4, which was not the right answer. Then I saw it said distance from explosion, so it would be the hypotenuse of a 1.22 x 10^4 by 2700 triangle, giving 1.25 x 10^4.. which was still not the right answer. Anyone see where I may have gone wrong?
 
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I don't see any mistakes. What did they give as the answer?
 
I didn't check your work on the pythagorean theorem. Otherwise, everything else looked fine to me... I'm scratching my head as well, except that the rocket sounds a bit heavy...

If you know the correct answer, perhaps the problem can be worked backward?
 

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