Ultimately, the bullet hits the same target regardless of the motion of the ship.
Observer on the ship perceives no difference in speed or direction.
Observer outside the ship sees bullet move at .6614 c.
So the speed is not 0.7.
There are a lot of formulas in that link and it is hard to follow, and some if what is written there is incorrect.
Can you tell me the exact formula to determine speed of two velocities, both 0.5C, one on the x and one on the y axis?
In any case, the impact location...
One last question, and as far as my purposes, the thread can be closed.
Is there any way for the guy on the train to determine he is moving, assuming he cannot see outside of the train?
Awesome - got it it just clicked.
Other thing I was worried about was the need for a Lorentz transformation here - but it doesn't apply, as we remain in the same reference frame, therefore the Pythagorean theorem holds for this specific example.
Okay - a bullet moving 0.5c north inside a train also going 0.5c east would be seen by an outside of the train observer at 0.707c northeast.
For an observer on the train - bullet still perceived at 0.5 c, and despite the added motion, the bullet would still strike the same target on the train...
Thanks.
So, the observer on the ship still sees the bullet moving at 0.5c, even though the ship is moving 0.3. Addition of velocities keeps it on target and same speed relative to observer on ship, but observer outside sees it diagonally higher speed.
Good evening,
If a ship is moving at 0.5C east in a gravity free environment, and fires a bullet perpendicular at 0.5c, is it just the hypotenuse (a squared + b squared)/c squared that we use for the speed?
Or is there more to it since we are at great speed?