A $300 \mathrm{~m}$ high cliff is where two water droplets fall freely one after the other. The first has already fallen $\frac{1}{1000} \mathrm{~mm}$ when the second begins its fall.

How far apart will the two droplets be at the moment when the first one reaches the base of the cliff? (The result should be calculated to the nearest $\frac{1}{10} \mathrm{~mm}$. Air resistance, etc., should not be considered.)
Let generally $300000 \mathrm{~mm}=s, \frac{1}{1000} \mathrm{~mm}=s_{1}$. The first water droplet falls for $t=\sqrt{\frac{2 s}{g}}$ time, while the second one starts its fall $t_{1}=\sqrt{\frac{2 s_{1}}{g}}$ time later, and thus the second water droplet falls for $t-t_{1}$ time until the first one reaches the base of the cliff. Therefore, the distance $d$ in question can be obtained by subtracting the distance $x$ that the second water droplet travels in $t-t_{1}$ time from $s$. Hence:

$$
x=\frac{1}{2} g\left(t-t_{1}\right)^{2}=\frac{1}{2} g\left(\sqrt{\frac{2 s}{g}}-\sqrt{\frac{2 s_{1}}{g}}\right)^{2}=\left(\sqrt{s}-\sqrt{s_{1}}\right)^{2}=s-2 \sqrt{s s_{1}}+s_{1}
$$

Thus,

$$
d=s-\left(s-2 \sqrt{s s_{1}}+s_{1}\right)=2 \sqrt{s s_{1}}-s_{1}
$$

Since we want to calculate the result to an accuracy of $\frac{1}{10} \mathrm{~mm}$, the subtracted term can be neglected, and thus

$$
d=2 \sqrt{s s_{1}}=20 \sqrt{3}=34.6 \mathrm{~mm}
$$

(Wohlstein Sándor, Érsekújvár.)

The problem was also solved by: Bartók I., Bayer B., Bogdán G., Holzer S., Izsáky L., Kalmár S., König D., Lázár L., Losonczy I., Messik G., Pazsiezky G., Picker G., Pilczer P., Pintér M., Pivnyik I., Póka Gy., Preisich Gy., Riesz K., Riesz M., Schlesinger A., Schmidl I., Sümegi Gy., Szmodics K., Tóbiás L., Weisz P.