= Coriolis deflection of a vertical projectile
{c}
{title2=$\Delta x_E=-\frac{4\omega v^3}{3g^2}\sin\theta$}
At colatitude $\theta$, ignore terms quadratic in Earth angular speed and use uniform vertical gravity. A vertically launched particle has zeroth-order height $z=vt-gt^2/2$. 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.
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