We throw a ball with an initial velocity $v_{0}$ downward from a height $H$ onto a horizontal surface, and after the collision, it rises again to a height $H$. After how much time will the object be at rest? What was $v_{0}$, if we also know that after the second collision, the ball rises to a height $H^{\prime}$? (Neglect air resistance!)
A ball thrown downward at a speed of $v_{0}$ reaches the ground at a speed of $v_{1}=\sqrt{2 g H+v_{0}^{2}}$. Since after the first collision it rises to a height of $H$, the speed after the collision is $v_{2}=\sqrt{2 g H}$. The coefficient of restitution is

$$
\varepsilon=\frac{v_{2}}{v_{1}}=\sqrt{\frac{2 g H}{2 g H+v_{0}^{2}}}
$$

The coefficient of restitution is constant, so for any two consecutive collisions, $v_{k+1}=\varepsilon v_{k}$, that is,

$$
v_{k+1}=\varepsilon^{k-1} \cdot v_{2} \quad(k=1,2,3, \ldots)
$$

The ball performs uniformly accelerated motion on each segment, so the time until the first collision is:

$$
t_{1}=\frac{v_{1}-v_{0}}{g}=\frac{\sqrt{2 g H+v_{0}^{2}}-v_{0}}{g}
$$

The time between the $k$-th and $(k+1)$-th collisions is twice the time of a vertical throw:

$$
t_{k+1}=2 \frac{u_{k+1}}{g}=\frac{2 v_{2}}{g} \varepsilon^{k-1}
$$

The total time until the $(k+1)$-th collision is therefore

$$
T_{k+1}=t_{1}+t_{2}+t_{3}+\ldots+t_{k+1}=t_{1}+\frac{2 v_{2}}{g}\left(\varepsilon^{0}+\varepsilon^{1}+\varepsilon^{2}+\ldots+\varepsilon^{k-1}\right)
$$

A certain fraction of the speed remains after each collision, so the ball can theoretically bounce many times. However, the infinite number of collisions occurs in a finite time, because (1) implies $\varepsilon<1$, and the sum of an infinite geometric series is finite:

$$
\varepsilon^{0}+\varepsilon^{1}+\varepsilon^{2}+\ldots+\varepsilon^{k-1}+\ldots=\frac{1}{1-\varepsilon}
$$

The required time is therefore

$$
T=t_{1}+\frac{2 v_{2}}{g} \frac{1}{1-\varepsilon}
$$

Substituting the previously expressed values:

$$
T=\frac{\sqrt{2 g H+v_{0}^{2}}-v_{0}}{g}+2 \sqrt{\frac{2 H}{g}} \frac{1}{1-\sqrt{\frac{2 g H}{2 g H+v_{0}^{2}}}}
$$

We know that after the second collision, the ball rises to a height of $H^{\prime}$, and therefore $v_{3}=\sqrt{2 g H^{\prime}}$. The coefficient of restitution can be written in two ways based on this:

$$
\varepsilon=\frac{v_{2}}{v_{1}}=\frac{v_{3}}{v_{2}}
$$

From this,

$$
\begin{aligned}
& \frac{\sqrt{2 g H}}{\sqrt{2 g H+v_{0}^{2}}}=\frac{\sqrt{2 g H^{\prime}}}{\sqrt{2 g H}} \\
& v_{0}=\sqrt{2 g H\left(\frac{H}{H^{\prime}}-1\right)}
\end{aligned}
$$

Eva Pálvölgyi (Bp., Madách I. Gym., III. o. t. ) Csaba Gáspár (Dunaújváros, Münnich F. Gym., III. o. t. )

Note. The actual course of the motion should not be imagined as the body bouncing infinitely. The heights of the rises decrease geometrically (with a ratio of $\varepsilon^{2}$), so after a certain time, the calculated height of the rise is comparable to the deformation of the ball. At this point, no actual rise occurs, and we cannot speak of a collision.

Since the above phenomenon only occurs after a large number of collisions, the calculated value of $T$ does not differ much from the actual value. The error resulting from this is generally much smaller than other errors (e.g., the error due to the inaccuracy of the gravitational acceleration).