### Inertial and noninertial reference frames

see Fig.1.

If we define the axis and $\theta$ as in Fig. 1, then we can write the velocity vector of the river as:
$\vec{v}=v\sin{\theta}\hat{y}-v\cos{\theta}\hat{z}$
since $\vec{\omega}=\omega\hat{z}$, we have that the coriolis correction to the acceleration vector is

$\vec{a_c}=-2\vec{\omega}\times\vec{v}=2\omega v \sin{\theta}\hat{x}$

Now observe at Fig. 2.

it shows a cross-section of the river, as it flows towards you. convince yourself that the new x' axis, which is pointing towards the left bank of the river, is the same as the previous x axis, so the river feels coriolis' acceleration
$a_{x'}=2\omega v \sin{\theta}$
also, it feels an acceleration downwards, we'll naturally call it g, which is comprised of the gravity acceleration and a small, insignificant, centrifugal correction. so we have
$a_{y'}=g$
and now we'll argue that
$h/L=a_x/a_y=\frac{2\omega v\sin{\theta}}{g}$