 Solution

Use local east, north and upward unit vectors at the equator, with Earth's rotation $\mathbf{\Omega }={\Omega }_E\hat{\mathbf{n}}$. 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 $h$ isThe Coriolis acceleration is $-2\mathbf{\Omega }\times \mathbf{v}$. Since north crossed with downward points west, its negative points east:Integrating with $x_E(0)={\dot{x}}_E(0)=0$ gives the equatorial Coriolis deflection in free fall,For $h=200\mathrm{m}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and ${\Omega }_E=2\pi /(86164\mathrm{s})=7.2921\times 10^{-5}{\mathrm{s}}^{-1}$, $t_f=6.39\mathrm{s}$ andThis 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.