I am not going to use the numbers you gave me because it is my preference to do a problem in general; you can stick the numbers in. The speed of the space ship as seen by the space station I will take as , the speed of the laser in both frames is c, the speed of the rocket in the space ship frame I will take as (Note that I assume that the space ship is moving toward the space station; if it were moving away you would have to replace β by −β everywhere.) The speed of the rocket in the space station frame is obtained from the velocity addition formula:
.
In the space station frame, the distance which must be traveled I will take to be L (the distance to the ship when it fires. So it is now straightforward to calculate the times t (the time for the laser) and tR (the time for the rocket):
and
.
Next we find the times and as seen by the space ship. Two things have to be taken into consideration: the distance between the two is length-contracted, , and in the time (or ) the space station moves up a distance (or ) to “meet” the laser (or rocket). So, now,
;
and
.