1. A cosmic-ray proton streaks through the lab with velocity 0.85c at an angle of 50o with the +x direction (in the xy plane of the lab). Compute the magnitude and direction of the proton’s velocity when viewed from frame S’ moving with β=0.72 in the +x direction. 2. Ux'= Ux-v/1-vUx/c^2 Uy'=Uy/[tex]\gamma(1-vUx/c^2)[/tex] 3. My attempt at the solution was to simply in an example of 1-D motion. From considering that a proton is moving at 0.85c in the positive x-direction (ignoring the angle), this can be compared when viewed from the S' frame, moving relative to S with 0.72c in the +x direction. The observer in S' measures the proton to be approaching in the +x-direction at 0.85c. Using the first equation U1x' = 0.85c - 0.72c/1-(0.85c)(0.72c)/c^2 = .34c. Introducing a y component makes the problem much more challenging and I'm not quite sure what to do. Do I work out the x and y directions independently? In which case, would the velocity for the y-direction be 0.85c*sin(50 degrees) and the x direction be 0.85c*cos(50 degrees). How would this appear to an observer in S'? That is, how would I calculate the direction as observed in S' and the magnitude? Thank you.