Projectile Motion of a fired rocket

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Homework Statement


(Q) A rocket is initially at rest on the ground. When its engines fire, the rocket flies off in a straight line at an angle 53.1 degrees above the horizontal with a constant acceleration of magnitude g. The engines stop at a time T after launch, after which the rocket is in projectile motion. You can ignore air resistance and that g is independent of altitude.

Find the maximum altitude reached by the rocket as well as the horizontal distance from the launch point to the point where the rocket hits the ground.



Homework Equations


s=1/2at^2, Hmax for projectile = u^2Sin^2[x]/2g, Time of flight of projectile = 2uSin[x]/g. Also, by equation of motion, u^2 = 2as


The Attempt at a Solution



For the Maximum height, it will be the height traveled in time T plus the maximum height reached when in projectile motion.

Therefore, H = 1/2(gSin[53.1])T^2 + u^2Sin^2[53.1]/2g
But u^2 = 2(gSin[53.1])(1/2(gSin[53.1])T^2).

However, the answer is supposed to be (18/25)gT^2.

As for the horizontal distance, I am completely clueless.

Please help me!!!

Thank-you very much for your time and effort.
 

Answers and Replies

  • #2
D H
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Use sin(53.1 degrees) = 0.8 and cos(53.1 degrees) = 0.6.
 

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