Let $\tau$ be an arbitrary time interval. A freely falling body falls for $\frac{\tau}{2}$, and the distances traveled in the subsequent whole $\tau$ time intervals are in what ratio to each other?
I. solution. Let the distances covered in the time intervals $\frac{\tau}{2}$ and the subsequent whole $\tau$ be: $s, s_{1}, s_{2}, s_{3}, \ldots$ etc.; then

$$
\begin{gathered}
s=\frac{1}{2} g\left(\frac{\tau}{2}\right)^{2} \\
s_{1}=\frac{1}{2} g\left(\frac{3 \tau}{2}\right)^{2}-\frac{1}{2} g\left(\frac{\tau}{2}\right)^{2}=g \tau^{2} \\
s_{2}=\frac{1}{2} g\left(\frac{5 \tau}{2}\right)^{2}-\frac{1}{2} g\left(\frac{3 \tau}{2}\right)^{2}=2 g \tau^{2} \\
s_{3}=\frac{1}{2} g\left(\frac{7 \tau}{2}\right)^{2}-\frac{1}{2} g\left(\frac{5 \tau}{2}\right)^{2}=3 g \tau^{2}
\end{gathered}
$$

thus

$$
s_{1}: s_{2}: s_{3}: \ldots=1: 2: 3: \ldots
$$

that is, as the consecutive whole numbers.

(István Kubinyi, Nagyszombat.)

II. solution.

\[

\]

Similarly, the average speed in the second $\tau$ time interval:

$$
\frac{1}{2}\left(\frac{3}{2} g \tau+\frac{5}{2} g \tau\right)=2 g \tau
$$

the distance covered in the second $\tau$ time interval:

$$
s_{2}=2 g \tau \cdot \tau=g \tau^{2}
$$

the average speed in the third $\tau$ time interval:

$$
\frac{1}{2}\left(\frac{5}{2} g \tau+\frac{7}{2} g \tau\right)=3 g \tau
$$

the distance covered in the third $\tau$ time interval:

$$
s_{3}=3 g \tau \cdot \tau=3 g \tau^{2}
$$

thus

$$
s_{1}: s_{2}: s_{3}: \ldots=1: 2: 3: \ldots
$$

(Sándor Erdélyi, Budapest.)