= Solution
Use a local horizontal plane, constant gravity and first order in Earth's angular <speed>. Let $x$ point along the initial horizontal shot, $y$ to the gunner's left and $z$ upward. At the North pole, $\boldsymbol\Omega=\Omega\hat{\mathbf z}$. Maximum unperturbed range at equal launch and landing heights requires a $45^\circ$ elevation, giving
$$
v_x=\frac{u}{\sqrt2},\qquad t_f=\frac{\sqrt2u}{g},\qquad R_0=\frac{u^2}{g}.
$$
The leading <Coriolis acceleration> is $\ddot y=-2\Omega v_x$, while vertical motion has no first-order Coriolis correction. With zero initial lateral <displacement> and <velocity>,
$$
y(t_f)=-\Omega v_x t_f^2,\qquad
\boxed{|y(t_f)|=\frac{\sqrt2\Omega u^3}{g^2}\simeq1.34\times10^2\,\mathrm m.}
$$
Here $u=500\,\mathrm{m\,s^{-1}}$, $g\simeq9.81\,\mathrm{m\,s^{-2}}$ and $\Omega\simeq7.2921\times10^{-5}\,\mathrm{s^{-1}}$. The miss is \b[to the right of the aiming direction], independent of its azimuth at the pole. The flight time is about $72.1\,\mathrm s$, so $\Omega t_f\simeq0.0053\ll1$. This <polar projectile Coriolis deflection> uses the usual flat, uniform-gravity approximation; planetary <curvature> and variation of gravity are beyond its accuracy. Using the polar gravity rather than $9.81$ changes the estimate by less than one percent. Earth's sidereal rotation period, rather than the solar day, sets $\Omega$.<Earth Fact Sheet by National Aeronautics and Space Administration>[Orbital parameters: sidereal rotation period 23.9345 hours; bulk parameters: polar surface acceleration]{ref}
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2011/paper-1-polar-projectile.png]
{title=Polar projectile deflection}
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