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B ______________ A
C
I throw light from A to B, reflect it to C. Similarly I throw a ball at 0.1c at same time.
Now I see from C. Measure distance from A to C and time it took to reach me. It's speed was c. Now , if I compare ball's speed to that of light while standing at C, will I find it slower than 0.1c?
If I relate two speeds while standing at C , what will be their comparative speeds?
In your stationary system the ball and the photon will have a velocity difference of
$$\text{c}-\text{v}=0.9\text{ c}$$
In the moving system of the ball it's relative velocity to the photon is
$$\frac{\text{c}-\text{v}}{1-\frac{\text{c v}}{\text{c}^2}} = \text{c}$$
because the speed of light is the same in every inertial frame of reference (see Einstein's velocity addition formula).
If you want to use the formula in 2 or 3 spatial dimensions you have to split into parallel and transversal components:
$$v_{\text{x}}=\frac{\text{v}_{\text{x}}-\text{v}}{1-\frac{\text{v}_{\text{x}} \text{v}}{\text{c}^2}}$$
for the parallel and
$$v_{\text{y}}=\frac{\text{v}_{\text{y}}\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}{1-\frac{\text{v}_{\text{x}} \text{v}}{\text{c}^2}}$$
for the transversal component:
where the $\text{v}$s are the velocities in the stationary (relative to A, B, C) system and the $v$s for the velocities in the system that moves with $\text{v}$.
So if the ball is on it's way from A to B while the photon is already on it's way from B to C the relative velocity of the photon in the system of the ball is with Pythagoras
$$v=\sqrt{v_{\text{x}}^2+v_{\text{y}}^2}$$
which is still $\text{c}$ in your scenario where
$$\text{v}=0.1\text{ c}, \ \text{v}_{\text{x}} = \frac{\text{c}}{\sqrt{2}}, \ \text{v}_{\text{y}} = \frac{\text{c}}{\sqrt{2}}$$
when we assume a 45° angle.
You will calculate that the ball is travelling at 10% of the speed of light.
