= Solution
A <fictitious force> is a term required when writing <Newton's second law> in an accelerating frame; it comes from the coordinate choice rather than a new interaction. For axes rotating at constant $\boldsymbol\Omega$ about a fixed inertial origin,
$$
\left(\frac{d\mathbf A}{dt}\right)_{\!I}
=\left(\frac{d\mathbf A}{dt}\right)_{\!R}+\boldsymbol\Omega\times\mathbf A.
$$
Apply this identity twice to position to get $\mathbf a_I=\mathbf a_R+2\boldsymbol\Omega\times\mathbf v_R+\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf r)$. Moving the last two terms to the right of $m\mathbf a_I=\mathbf F$ gives the stated Coriolis and <centrifugal forces>. An accelerating origin or changing rotation rate would add further terms.
A slowly rotating fluid planet follows an equipotential of gravitational plus <centrifugal potential>. The <centrifugal potential> is $-\Omega^2r^2\cos^2\lambda/2$, most negative at the equator. An outward equatorial shift raises <gravitational potential> and restores the balance, producing an oblate figure. Self-gravity also changes with the shape; no flattening coefficient follows from rotation alone without a <density> model.
Using local north and radial-up <unit vectors>, the <centrifugal acceleration> on a stationary <mass> is
$$
\mathbf a_{\rm cf}=\Omega^2R\big(\cos^2\lambda\,\mathbf e_U
-\sin\lambda\cos\lambda\,\mathbf e_N\big).
$$
It points away from the rotation axis: upward and equatorward, at angle $\lambda$ to radial up in the northern hemisphere. Its vertical fraction of gravitational <force> is
$$
\boxed{\frac{F_{{\rm cf},U}}{mg}=\frac{\Omega^2R}{g}\cos^2\lambda
\simeq3.45\times10^{-3}\cos^2\lambda,}
$$
using a $24\,\mathrm h$ day for this approximate number. The resultant effective weight tilts equatorward relative to radial down. A plumb-defined local vertical already follows effective gravity; the component calculation uses the spherical radial vertical. A supported stationary <mass> of course has zero net <acceleration>.
For the <Coriolis deflection of vertical launch and drop>, retain only first order in $\Omega$, use effective constant $g$, and neglect horizontal <velocities> in the Coriolis term as instructed. With east positive, vertical <velocity> gives $\ddot x=-2\Omega\cos\lambda\,\dot z$ and no northward <acceleration>. For launch, $z=ut-gt^2/2$ and the flight time is $2u/g$. Integrating with zero initial horizontal <velocity> gives
$$
\dot x=-2\Omega\cos\lambda\,z,\qquad
\boxed{x_X=-\frac{4\Omega u^3\cos\lambda}{3g^2},\quad y_X=0.}
$$
The landing is westward. The westward <velocity> accumulated on ascent is only recovered on descent, so the two displacement contributions do not cancel.
For release from rest at $h_0=u^2/(2g)$, $z=h_0-gt^2/2$, $t_{\rm fall}=u/g$, and $\ddot x=2\Omega g t\cos\lambda$. Thus
$$
\boxed{x_{X'}=\frac{\Omega u^3\cos\lambda}{3g^2}=-\frac14x_X,\quad y_{X'}=0.}
$$
The drop lands eastward with one quarter the launch-return displacement magnitude. North–south corrections from horizontal Coriolis coupling are beyond the retained approximation.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2013/paper-1-rotating-earth.png]
{title=Original centrifugal geometry and opposite east–west deflections for launch-return and drop trajectories}
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