Projectile Motion of a fired rocket

[mit_hacker]

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.

 

[D H]

Use sin(53.1 degrees) = 0.8 and cos(53.1 degrees) = 0.6.

 

[m2003]

Dear student,

It seems like you are on the right track with your solution. I would recommend breaking down the problem into two parts: the rocket's motion while the engines are firing, and its motion after the engines stop.

First, let's focus on the rocket's motion while the engines are firing. We can use the equations of motion to determine its velocity and displacement at any given time. Since the acceleration is constant and the rocket starts from rest, we can use the equation v = u + at to find the rocket's velocity at time T (when the engines stop). We know that the rocket is traveling at an angle of 53.1 degrees above the horizontal, so we can use the components of this velocity to find its horizontal and vertical components. The horizontal component will remain constant after the engines stop, while the vertical component will decrease due to gravity.

Next, we can use the equations of projectile motion to determine the rocket's maximum height. We know that the rocket will reach its maximum height when its vertical velocity is equal to zero. We can use the equation v^2 = u^2 + 2as, where v is the final vertical velocity (zero), u is the initial vertical velocity (from the engines), a is the acceleration due to gravity, and s is the displacement (maximum height). Solving for s, we get s = u^2/2a. We can substitute our values for u and a to get s = (gSin[53.1])^2/2g, which simplifies to (18/25)g.

For the horizontal distance, we can use the equation s = ut + 1/2at^2, where s is the horizontal displacement, u is the initial horizontal velocity (from the engines), a is the acceleration (zero after the engines stop), and t is the time (T). We know that the rocket's initial horizontal velocity is the same as its final horizontal velocity, so we can use the same value we found earlier. Substituting our values, we get s = (gCos[53.1])T, which simplifies to (7/25)gT.

I hope this helps! Let me know if you have any further questions. Keep up the good work in your studies!