# Hard Modern Physics Problem

1. Jun 13, 2013

### Dgonzo15

1. The problem statement, all variables and given/known data
Persons A and B are rowing boats due east. B at .5c, and A at .7c, both relative to the earth, and are in parallel tracks 3 light seconds apart (B is north of A). A starts out behind B, but as A passes B, A throws a stopwatch straight up at .5c in A's reference frame. B fires a cannonball due south at a 45 degree angle to the horizon and it strikes the stop watch. If B fired the cannon the same time A threw the stopwatch, what is the muzzle velocity of the cannon?

2. Relevant equations
u'(x)=(u-v)/(1-(uv)/c^2)
u'(y)=u(y)/(γ(1-(u(x)v)/c^2))
γ=1/√(1-(v^2/c^2))

3. The attempt at a solution

WHAT I did first was convert the speed at which the watch was thrown relative to B (I did not include horizontal velocity in the first try, assuming the cannon was shot STRAIGHT south). After I got the vertical velocity of the watch in B's R.F (√(3)c/4), I used simple trig (given the 3 light seconds apart and the watch's velocity) to find a velocity for the cannon of .612c.
This was the wrong answer, so in my second attempt,

I included the horizontal velocity of the stopwatch as well (because A is moving, and I got 4c/13). With that, I again used some trig with multiple triangles to get the velocity of the cannonball (which, obviously couldn't have pointed EXACTLY south due to the horizontal motion of the watch) and got an answer of .685c, which is still wrong. Can anybody help??

2. Jun 13, 2013

### Staff: Mentor

What do you get as new vertical velocity of the stopwatch (as seen by B), if you take the motion of A into account?

Once you know this (together with the correct 4c/13 horizontal motion), the muzzle velocity is just 3D-geometry in B's system.