Relativity: Projectile Motion at an Angle on Spaceship

[mochi_melon]

Homework Statement

On a spaceship moving 5/6C, an astronaut fires a light pulse from a gun. The barrel of the gun makes an angle of 60 degrees from the X direction relative to the spaceship. What angle does the gun make in the rest frame of an observer on Earth and what angle does the light pulse move for the same Earth observer?

Homework Equations

u^{'}_{x}=C*cos\theta
V^{'}_{x}=5/6C
u_{x}=\stackrel{u^{'}_{x}+V^{'}_{x}}{1+\stackrel{u^{'}_{x}V^{'}_{x}}{c^{2}}}

For some reason the formatting didn't work out - the last equation is Einstein's Relative Addition formula.

The Attempt at a Solution

I found a value of 2.8216x10^{8} for the relative velocity in the x direction of the light pulse to the Earth observer, but I do not know what the next step would be. I suspect that the angle of the gun itself does not change, but I'm not sure how to prove or disprove this. Thanks!

 

[mochi_melon]

Ok, upon looking in my notes once more I found how to find the velocity of the light pulse as measured by the Earth observer and thus its components/angle, but I am still at a total loss to find at what angle the gun itself is in the stationary reference frame. Since that was the part A I suspect it is something very simple I am overlooking :(

 

[robphy]

With the gun as the hypotenuse, draw a right-triangle with legs along the x- and y-axes.
What is the corresponding triangle in the Earth's frame?

 

[mochi_melon]

I wound up following the formulas (copied from another website since I'm having trouble having the division lines come up properly for these formulas)

wx = (ux + vx) / (1 + ux vx / c2) = (48/51)c
wy = uy / [(1 + ux vx / c2) gamma(vx)] = .337915c
wz = uz / [(1 + ux vx / c2) gamma(vx)] = 0

Where w is the velocity seen on Earth, u is the pulse velocity and v is the spaceship. I got an angle that is way less than the original (19.75^{o}) Does that make sense?


As for the angle of the gun itself, I used these equations:
L_{x}=\frac{l_{ox}}{\gamma(v_{x})}=l_{o}cos60\sqrt{11/36}
L_{y}=\frac{l_{oy}}{\gamma(v_{y})}=l_{o}sin60

Resulting in an angle of 72.3^{o}. Again, is this a logical answer or did I do something wrong?