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Take $0\le \gamma \le 1$ for the coefficient of restitution. Conservation of energy in the initial vacuum fall givesThe 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 ${\gamma }^2$; henceGravity 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 hasThe total travelled distance for $\gamma <1$ is the first fall plus ascent and descent for every rebound. Summing the geometric series givesThe 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 givesBefore impact $u\le 0$, so $\dot{u}=-g+(\alpha /m)u^2$. Set $v=-u\ge 0$, $v_T=\sqrt{mg/\alpha }$, and $\beta =\sqrt{\alpha g/m}=g/v_T$. Separation gives $\mathrm{artanh}(v/v_T)=\beta t$, establishing the descent from rest under quadratic drag:The speed approaches the terminal velocity $v_T$, although the formula is used only until the first impact. This occurs when $\cosh (\beta t_{\mathrm{hit}})=e^{\alpha h_0/m}$. 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 $\alpha \to 0$.