= Solution
Take $0\le\gamma\le1$ for the <coefficient of restitution>. <Conservation of energy> in the initial vacuum fall gives
$$
\boxed{u_0=\sqrt{2gh_0}.}
$$
The fixed plate has zero velocity, so the restitution rule makes the first upward rebound speed $u_1=\gamma u_0$. The ratio of rebound to impact <kinetic energy> is $\gamma^2$; hence
$$
\boxed{u_1=\gamma\sqrt{2gh_0},\qquad\text{fraction dissipated}=1-\gamma^2.}
$$
Gravity preserves the speed magnitude between departure from and return to the same plate height. Therefore the upward speed after the $n$th bounce is $\gamma^n u_0$. The <geometric bounce sequence> has
$$
\boxed{h_n=\gamma^{2n}h_0,\qquad T_n=\frac{2\gamma^n u_0}{g}=2\gamma^n\sqrt{\frac{2h_0}{g}}\quad(n\ge1).}
$$
The total travelled distance for $\gamma<1$ is the first fall plus ascent and descent for every rebound. Summing the <geometric series> gives
$$
\boxed{D=h_0+2\sum_{n=1}^{\infty}h_0\gamma^{2n}
=h_0\frac{1+\gamma^2}{1-\gamma^2}.}
$$
The idealized infinitely many impacts accumulate in finite time, after which the limiting state is rest on the plate; $\gamma=0$ gives rest at the first impact.
For the atmospheric fall, use upward signed velocity $u$ and height $z$. <Newtonian mechanics> with <quadratic drag> gives
$$
m\dot u=-mg-\alpha|u|u,\qquad u(0)=0,\qquad z(0)=h_0.
$$
Before impact $u\le0$, so $\dot u=-g+(\alpha/m)u^2$. Set $v=-u\ge0$, $v_T=\sqrt{mg/\alpha}$, and $\beta=\sqrt{\alpha g/m}=g/v_T$. Separation gives $\operatorname{artanh}(v/v_T)=\beta t$, establishing the <descent from rest under quadratic drag>:
$$
\boxed{u(t)=-\sqrt{\frac{mg}{\alpha}}\tanh\!\left(\sqrt{\frac{\alpha g}{m}}\,t\right),\qquad
z(t)=h_0-\frac m\alpha\log\cosh(\beta t).}
$$
The speed approaches the <terminal velocity> $v_T$, although the formula is used only until the first impact. This occurs when $\cosh(\beta t_{{\rm hit}})=e^{\alpha h_0/m}$. In particular, the atmospheric impact speed is $v_T\sqrt{1-e^{-2\alpha h_0/m}}$, and both speed and height formulas recover the vacuum fall as $\alpha\to0$.
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