# Vertical path of the michelson-morley experiment setup

In an exercise I am asked to calculate the time taken for the light to travel the path $ct=\sqrt{L^2-v(\frac{t}{2})^2}\Rightarrow t=2\frac{L}{\sqrt{c^2-v^2}}$. Where the Pythagorean theorem was used to calculate the slanted path the light must travel with $L$ being the length of the vertical path from the first mirror (at the intersection paths) to the second mirror and $vt$ being the distance earth has traveled until the light reaches the second mirror. Now what I don't understand is this: why can we equate the distance of the vertical path(s) to the slanted path(s) with $ct$ being the vertical path since $c$ is only the speed the light is traveling in the vertical path? Isn't the horizontal component of the light's velocity still $v$ if we assume or don't assume the existence of ether?