= Solution
As the real charge descends to height $z(t)$, its instantaneous <image charge> is at $-z(t)$. The force is therefore
$$
m\ddot z=-\frac{q^2}{4\pi\varepsilon_0(2z)^2},\qquad \ddot z=-\frac{\kappa}{z^2},\qquad \kappa=\frac{q^2}{16\pi\varepsilon_0m}.
$$
The image is not kept at its initial position. Multiplying by $\dot z$ and using $z(0)=d$, $\dot z(0)=0$ gives
$$
\frac12\dot z^2=\kappa\left(\frac1z-\frac1d\right),\qquad \dot z<0.
$$
For the <falling charge above a grounded conducting plane>, the arrival time is
$$
T=\frac1{\sqrt{2\kappa}}\int_0^d\frac{dz}{\sqrt{1/z-1/d}}.
$$
Put $z=d\sin^2\theta$; the integral becomes $2d^{3/2}\int_0^{\pi/2}\sin^2\theta\,d\theta=\pi d^{3/2}/2$. Hence
$$
\boxed{T=\frac{\pi d^{3/2}}{2\sqrt{2\kappa}}=\frac\pi{|q|}\sqrt{2\pi\varepsilon_0md^3}.}
$$
This is the finite arrival time in the specified ideal force model, for $q\ne0$. The field-force derivation also avoids incorrectly using the full real-image pair interaction energy: the conductor's effective energy is $-q^2/(16\pi\varepsilon_0z)$, with the factor one-half appropriate to induced charge.
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