= 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> $2\mathbf v\times\boldsymbol\Omega_E$ points east and has magnitude $2\Omega_Egt$. Integrating from zero initial eastward velocity and displacement gives
$$
v_x=\Omega_Egt^2,\qquad x=\frac{\Omega_Eg t^3}{3},\qquad t_f=\sqrt{2h/g}.
$$
For $h=830\,\mathrm m$, $g\simeq9.81\,\mathrm{m\,s^{-2}}$ and $\Omega_E\simeq7.29\times10^{-5}\,\mathrm{s^{-1}}$,
$$
\boxed{t_f\simeq13.0\,\mathrm s,\qquad x_f\simeq0.525\,\mathrm m\ \text{east of the worker}.}
$$
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 $+\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf r)$, its sign is opposite the usual outward inertial <force>; it does not affect the retained first-order Coriolis result.
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