= Solution
At instantaneous height $z$, the <image charge> is a distance $2z$ below the real charge, giving force $m\ddot z=-K/z^2$ with $K=q^2/(16\pi\varepsilon_0)$. Multiplying by $\dot z$ and using rest at $z=d$ gives
$$
\frac m2\dot z^2=K\left(\frac1z-\frac1d\right).
$$
The induced-conductor potential energy is $-K/z$, not the full energy of a pair of independently movable charges. On the falling branch, integrate the reciprocal speed:
$$
t_{\rm hit}=\sqrt{\frac m{2K}}\int_0^d\sqrt{\frac{zd}{d-z}}\,dz=\sqrt{\frac{md^3}{2K}}\int_0^1\sqrt{\frac{s}{1-s}}\,ds=\frac\pi2\sqrt{\frac{md^3}{2K}}.
$$
The last <integral> is $\pi/2$ by $s=\sin^2\theta$. Hence the <fall time of a charge toward a grounded plane> in the electrostatic Newtonian model is
$$
\boxed{t_{\rm hit}=\frac{\pi\sqrt{2\pi\varepsilon_0md^3}}{|q|}.}
$$
For $q=0$ there is no force and the particle remains at rest.
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