= Equatorial Coriolis deflection in free fall
{title2=$x_E=\Omega_E(2h)^{3/2}/(3\sqrt g)$}
A body released from rest relative to <Earth> at the equator has unperturbed downward <velocity> $v_U=-gt$. The <Coriolis acceleration in local Earth coordinates> gives eastward acceleration $a_E=-2\Omega_Ev_U=2\Omega_Egt$. Integration gives $x_E=\Omega_Egt^3/3$ and impact time $t_f=\sqrt{2h/g}$, proving the title formula. This is first order in <Earth's rotation>, neglecting air resistance and the height dependence of the effective local <gravitational acceleration>. For a short fall at latitude $\varphi$, its leading eastward term is multiplied by $\cos\varphi$.
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