 Fall time of a charge toward a grounded plane

The image of a charge $q$ at height $z$ is $-q$ at height $-z$. Their separation is $2z$, giving force $-q^2/(16\pi {\epsilon }_0{z}^2)$. Starting at rest at height $d$, energy conservation gives ${\dot{z}}^2=q^2(1/z-1/d)/(8\pi {\epsilon }_{0}m)$. Integrating the reciprocal speed to the plane yields the stated finite time in the electrostatic Newtonian model. The induced-conductor potential energy is $-q^2/(16\pi {\epsilon }_{0}z)$; using the full image-pair energy would double the force incorrectly.