Special relativity kinematics problem I have this problem in SR kinematics.
a spaceship travels near earth at c/2. it shoots off a light-ray at 45deg to its direction of travel (measured in its own ref frame). What is this angle in the earth's ref frame?
the answer is supposedly about 19 degrees, but there is no explanation. can anyone explain why?
the problem was posed in an exam at a college level two semester intro physics course for computer science (yeah, dont ask about the relevance, i have no clue either...) and it should be solvable with just length contraction, time dilation and addition of velocities (as these were the discussed topics)
 A: If we say that the spaceship and earth are very close to each other, let us then imagine that the spaceship, right as it passes by earth, shoots a photon out at $45^{\circ}$ from its direction of travel.  The simplest way to attack this problem will be with the transformation of velocities in the x-direction.  From our known information, we can determine that the photon's velocity in the x-direction in the spaceship's frame is: $$ux^1 = c *cos(\theta^1)$$   In the earth's frame, the photon is seen to move at a speed: $$ ux = c*cos(\theta)$$  From here we can plug into the velocity addition formula and solve for $\theta$.  
Velocity addition formula = 

From here:
$$ccos(\theta) =  \frac{ccos(\theta^1)+v}{1+\frac{v ccos(\theta^1)}{c^2}}$$
Divide by c:
$$cos(\theta) =  \frac{cos(\theta^1)+\frac{v}{c}}{1+\frac{v cos(\theta^1)}{c}}$$
$$cos(\theta) =  \frac{cos(\theta^1)+\beta}{1+\beta cos(\theta^1)}$$
From here, all that's left to do is plug in for beta and the angle from the spaceship and you can find $\theta$
