Coriolis deflection of a vertical projectile ($\Delta x_E=-\frac{4\omega v^3}{3g^2}\sin \theta$)

At colatitude $\theta$, ignore terms quadratic in Earth angular speed and use uniform vertical gravity. A vertically launched particle has zeroth-order height $z=vt-g{t}^2/2$. Its eastward Coriolis acceleration is $-2\omega \sin \theta \dot{z}$, so its horizontal velocity is $-2\omega \sin \theta z$. Integrating over the flight time $2v/g$ gives the westward displacement displayed above. Descent reverses the horizontal acceleration but does not cancel the accumulated displacement.