# Deviation of free falling objects (Coriolis effect) using conservation of angular momentum

I read this pdf on non inertial frame, in particular I have a question on the deviation of free falling object due to Coriolis effect.

Consider a ball let go from a tower at height $h$. The displacement due to Coriolis effect, calculated with formulas in Earth system, is $(4.19)$, after it there is explanation of the effect that uses the conservation of the angular momentum of the ball in a inertial frame.

$$x =\frac{2\sqrt{2}ωh^{3/2}}{3g^{1/2}} \tag{4.19}$$ Just before being dropped, the particle is at radius $(R+h)$ and co-rotating, so it has speed $(R+h)ω$ and angular momentum per unit mass $(R+h)^2ω$. As it falls, its angular momentum is conserved (the only force is central), so its ﬁnal speed v in the (Eastward) direction of rotation satisﬁes $Rv = (R+h)^2ω$, and $v= (R+h)^2ω/R$. Since this is larger than the speed $Rω$ of the foot of the tower, the particle gets ahead of the tower. The horizontal velocity relative to the tower is approximately $2hω$ (ignoring the $h^2$ term), so the average relative speed over the fall is about $hω$. We now see that the displacement $(4.19)$ can be expressed in the form (time of ﬂight) times (average relative velocity) as might be expected.

But $$v_{average} t_{flight}=h \omega \sqrt{\frac{2h}{g}}$$

Which differs by $\frac{2}{3}$ from $(4.19)$. Is that due to the approximation made?

I also don't understand completely why the average relative velocity $v_{average}$ is taken to be half the relative velocity found. Isn't this valid only for constant accelerated linear motions?

The error is just to consider an average speed $h\omega$.
When the particle is at height $z$, its horizontal (relative to the Earth) speed is $v=2z\omega$. The time of of flight is $$t=\sqrt{\frac{2z}{g}}.$$ Differentiating this expression we get the time taken by the particle to move a distance $dz$, $$dt=\frac{dz}{\sqrt{2gz}}.$$ The horizontal distance traveled during this $dt$ is $$dx=vdt=2z\omega\frac{dz}{\sqrt{2gz}}.$$ Integrating from $0$ to $h$ we obtain the total horizontal displacement $$x=\sqrt{\frac 2g}\omega\int_0^h\sqrt zdz=\frac{2\omega}{3}\sqrt{\frac{2h^3}{g}}.$$