How far does a body thrown at an angle $\alpha$ with an initial velocity $v_{0}$ jump, and how much distance does it cover on the horizontal ground during this time, if the coefficient of restitution is $\varepsilon$? Ignore air resistance!
The motion of a test consists of multiple consecutive oblique throws. The horizontal component of its initial velocity, $v_{0} \cdot \cos \alpha$, remains constant throughout the motion, while the vertical component, $v_{0} \sin \alpha$, decreases to $\varepsilon v_{0} \sin \alpha$ after the first bounce, $\varepsilon^{2} v_{0} \sin \alpha$ after the second bounce, and so on.

The time elapsed until the first collision is obtained from the condition that the vertical component of the test's velocity is 0 at half the time, i.e., at the apex of the parabolic trajectory:

$$
v_{0} \sin \alpha - g t_{1} / 2 = 0, \quad \text { hence } \quad t_{1} = \left(2 v_{0} / g\right) \sin \alpha
$$

The vertical component of the initial velocity for the next throw is $\varepsilon v_{0} \sin \alpha$, so the time required to complete this segment is

$$
t_{2} = \frac{2 \varepsilon v_{0}}{g} \sin \alpha \quad \text { time is needed. }
$$

Similarly, $t_{3} = \frac{2 \varepsilon^{2} v_{0}}{g} \sin \alpha, \ldots, t_{n} = \frac{2 \varepsilon^{n-1} v_{0}}{g} \sin \alpha$. The total time is

$$
\begin{aligned}
t = t_{1} + t_{2} + \ldots + t_{n} + \ldots & = \frac{2 v_{0}}{g} \sin \alpha \left(1 + \varepsilon + \varepsilon^{2} + \ldots + \varepsilon^{n-1} + \ldots\right) = \\
& = \frac{2 v_{0}}{g} (\sin \alpha) \frac{1}{1 - \varepsilon}
\end{aligned}
$$

During this time, the test travels horizontally

$$
s = \left(v_{0} \cos \alpha\right) \cdot t = \frac{2 v_{0}}{g (1 - \varepsilon)} (\sin \alpha) \cdot v_{0} \cdot \cos \alpha = \frac{v_{0}^{2} \sin 2 \alpha}{g (1 - \varepsilon)}
$$

After the bouncing ceases, the test performs uniform motion at a speed of $v_{0} \cos \alpha$.

Egon Szurmai (Ózd, József A. Gymnasium, 3rd grade)