= Solution
Neglect the <centrifugal acceleration>, which is of order $\omega^2$, and use the local flat-surface approximation. To zeroth order in $\omega$, the height above $P$ and flight time are
$$
z_0(t)=vt-\frac12gt^2,\qquad \dot z_0=v-gt,\qquad T_0=\frac{2v}{g}.
$$
Inserting this vertical <velocity> into the <Coriolis acceleration> gives the eastward component
$$
\ddot x=-2\omega\sin\theta\,(v-gt)+O(\omega^2).
$$
With $x(0)=\dot x(0)=0$, two integrations yield
$$
\dot x=-2\omega\sin\theta\left(vt-\frac12gt^2\right),\qquad
x=-\omega\sin\theta\left(vt^2-\frac13gt^3\right)
$$
to first order. Horizontal <velocity> is already of order $\omega$, so its effect on vertical acceleration is of order $\omega^2$. Thus the flight time remains $T_0$ at this order. The <Coriolis deflection of a vertical projectile> at landing is
$$
\boxed{x(T_0)=-\frac{4\omega v^3}{3g^2}\sin\theta.}
$$
The displacement is \b[westward], and its magnitude is $4\omega v^3\sin\theta/(3g^2)$. The horizontal speed remains westward during the flight: $\dot x=-2\omega\sin\theta\,z_0(t)$. This explains why the eastward acceleration on descent does not cancel the earlier displacement.
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