= 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 \b[$v_0=\sqrt{2qEh/m}$].
After the $n$th bounce, the outward speed is $\gamma^nv_0$, so the outward height is $\gamma^{2n}h$ and that excursion contributes twice this distance. Hence
$$
L=h+2h\sum_{n\geq1}\gamma^{2n}
=\boxed{h\frac{1+\gamma^2}{1-\gamma^2}},
$$
so $q_1(\gamma)=1+\gamma^2$ and $q_2(\gamma)=1-\gamma^2$.
With quadratic drag, put $w=-\dot z\geq0$. Then
$$
m\dot w=qE-\alpha w^2,\qquad w(0)=0,
$$
whose solution is $w=\sqrt{qE/\alpha}\tanh(\sqrt{\alpha qE}\,t/m)$. Integration gives
$$
\boxed{z(t)=h-\frac m\alpha\log\cosh\left(\frac{\sqrt{\alpha qE}}m\,t\right).}
$$
Thus $A=m/\alpha$, $B=\sqrt{\alpha qE}/m$, and $f(s)=\log\cosh s$.
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