= Solution
Use local east, north and upward <unit vectors> at the equator, with <Earth's rotation> $\boldsymbol\Omega=\Omega_E\widehat{\mathbf n}$. The parcel starts at rest relative to the ground. To first order in $\Omega_E$, its unperturbed vertical <velocity> is $\dot z=-gt$ and its fall time from height $h$ is
$$
t_f=\sqrt{\frac{2h}{g}}.
$$
The <Coriolis acceleration> is $-2\boldsymbol\Omega\times\mathbf v$. Since north crossed with downward points west, its negative points east:
$$
\ddot x_E=-2\Omega_E\dot z=2\Omega_Egt.
$$
Integrating with $x_E(0)=\dot x_E(0)=0$ gives the <equatorial Coriolis deflection in free fall>,
$$
x_E(t)=\frac13\Omega_Egt^3,\qquad
\boxed{x_E(t_f)=\frac{\Omega_E}{3}\frac{(2h)^{3/2}}{\sqrt g}.}
$$
For $h=200\ \mathrm m$, $g=9.81\ \mathrm{m\,s^{-2}}$ and $\Omega_E=2\pi/(86164\ \mathrm s)=7.2921\times10^{-5}\ \mathrm{s^{-1}}$, $t_f=6.39\ \mathrm s$ and
$$
\boxed{\text{miss distance}\simeq0.0621\ \mathrm m=6.2\ \mathrm{cm},\quad\text{eastward}.}
$$
This neglects air resistance, variation of $g$ over the fall, and terms of order $\Omega_E^2$. The measured local $g$ already includes the background centrifugal correction. “Stationary” is interpreted as hovering above a fixed ground location, not being stationary in an inertial frame.
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