 Solution

Use local east, north and upward axes. At the equator, Earth's angular velocity points north. To first order in its magnitude ${\Omega }_E$, the falling object has $v_z=-gt$, so the Coriolis acceleration $2\mathbf{v}\times {\mathbf{\Omega }}_E$ points east and has magnitude $2{\Omega }_{E}gt$. Integrating from zero initial eastward velocity and displacement givesFor $h=830\mathrm{m}$, $g\simeq 9.81\mathrm{m}{\mathrm{s}}^{-2}$ and ${\Omega }_E\simeq 7.29\times 10^{-5}{\mathrm{s}}^{-1}$,Local measured $g$ includes the stationary centrifugal correction. Corrections involving the acquired horizontal velocity or the height dependence of effective gravity are beyond this leading estimate. If the printed centrifugal term is read as $+\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})$, its sign is opposite the usual outward inertial force; it does not affect the retained first-order Coriolis result.