8.5 m deep lake we throw a stone, whose density is: 2.1. How long and with what speed does the stone reach the bottom of the lake:[1

${ }^{1}$ The acceleration due to free fall: $g=980.8 \frac{\mathrm{cm}}{\mathrm{sec}^{2}}$ (according to Gruber's determination in Budapest $g=980.837 \frac{\mathrm{cm}}{\mathrm{sec}^{2}}$ ).
We calculate the acceleration exhibited in water. The accelerations are proportional to the acting resultant forces, that is,

$$
g_{1}: g=p_{1}: p
$$

where $p$ and $g$ are the weight and acceleration in a vacuum, and $p_{1}$ and $g_{1}$ are in water. Considering Archimedes' principle,

$$
\begin{gathered}
p_{1}=v \sigma g-v \sigma_{1} g \\
p=v \sigma g
\end{gathered}
$$

where $v$ is the volume of the stone, $\sigma$ is its density, and $\sigma_{1}$ is the density of water, thus

$$
\begin{gathered}
g_{1}: g=\left(\sigma-\sigma_{1}\right): \sigma ; \quad g_{2}=\frac{1.1}{2.1} g=513.7 \frac{\mathrm{cm}}{\mathrm{sec}^{2}} \\
t=\sqrt{\frac{2 s}{g_{1}}}=1.82 \mathrm{sec} \\
v=g_{1} t=935 \frac{\mathrm{cm}}{\mathrm{sec}}
\end{gathered}
$$

(Csada Imre, Pápa.)

The problem was also solved by: Dömény I., Fodor H., Freund E., Fuchs I., Haar A., Jánosy Gy., Kiss J., Kräuter F., Merse P., Pető L., Rássy P., Rosenberg J., Schwarz Gy., Schwarz O., Székely J.