(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.
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.