 Solution

The Lorentz force law is $\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. Before impact the acceleration toward the wall is $a=qE/m$, so the first-impact speed is $v_0=\sqrt{2qEh/m}$.

After the $n$th bounce, the outward speed is ${\gamma }^n{v}_0$, so the outward height is ${\gamma }^{2n}h$ and that excursion contributes twice this distance. Henceso $q_1(\gamma )=1+{\gamma }^2$ and $q_2(\gamma )=1-{\gamma }^2$.

With quadratic drag, put $w=-\dot{z}\ge 0$. Thenwhose solution is $w=\sqrt{qE/\alpha }\tanh (\sqrt{\alpha qE}t/m)$. Integration givesThus $A=m/\alpha$, $B=\sqrt{\alpha qE}/m$, and $f(s)=\log \cosh s$.