Suppose a train is traveling past an observer at a velocity of v'. (Primes denote variables as measured by the ground observer; unprimed variables as measured by the observer on the train.) On that train a ball is thrown with a velocity of u and caught a distance Δx away after time interval Δt. The throwing and catching of the ball are witnessed andmeasured by the ground observer. The question is "What is the apparent velocity of the ball for the observer on the ground?" In Newtonian mechanics the answer is (v'+u), but Special Relativity has a different answer.

In the interval of time Δt on the train the ball traveled a distance uΔt. Because the train is moving with respect to the ground observer the distances on the train as observed by the ground observer are foreshortened and time intervals lengthened. For the ground observer the distance and time interval must be transformed using Lorentz' equations.The Lorentz transformations from the measurements on the train to measurements for the ground observer, who appears to be traveling at a velocity of −v', are

Δx' = γΔx − βγcΔt
Δt' = γΔt − βγΔx/c

where c is the speed of light, β=(−v')/c and γ=1/(1−β²)^½.

Now let Δx=uΔt and compute u'=(Δx'/Δt'), the apparent velocity of the ball with respct to the ground observer.

u' =Δx'/Δt/ = (γuΔt − βγcΔt)/(γΔt − βγuΔt/c)
where upon dividing
numerator and denominator
by γΔt gives
u' = (u − βc)/(1 − βu/c)
but βc is equal to −v'
so the above reduces to
u' = (v'+u)/(1+ uv'/c²)

This is the formula for the relativistic addition of velocities.

What a marvelous formula it is! Suppose v'=c and u=c. Their combined velocity is (2c)/(1+1)=c. Furthermore if u=c then the combined velocity is marvelously c no matter what v' is. That is as follows:

u' = (v'+c)/(1+v'c/c²)
or, equivalently
u' = (v'+c)/(1+v'/c)
where upon multiplying
numerator and denominator
by c gives
u' = c(v'+c)/(c+v') = c

This says that the speed of light is the same no matter the speed of the reference frame in which it is measured.