With what acceleration does a steel ball move into the sea, with what resultant force does it hit the seabed, and how long does its motion take? The depth of the sea is $m=100 \mathrm{~m}$. The diameter of the ball is $d=0.20 \mathrm{~m}$. The specific gravity of the steel ball is $\sigma=7.23$; the specific gravity of seawater is $\sigma_{1}=1.05$.
The weight of the ball: $\frac{d^{3} \pi}{6} \cdot \sigma$; the weight of the displaced water: $\frac{d^{3} \pi}{6} \cdot \sigma_{1}$; the weight of the ball in water: $\frac{d^{3} \pi}{6}\left(\sigma-\sigma_{1}\right)$. Since the mass of the ball is $\frac{d^{3} \pi}{6} \cdot \frac{\sigma}{g}$, the sought acceleration is:

$$
\gamma=\frac{d^{3} \pi}{6} \cdot\left(\sigma-\sigma_{1}\right): \frac{d^{3} \pi}{6} \cdot \frac{\sigma}{g}=\frac{\sigma-\sigma_{1}}{\sigma} g=8.38 \mathrm{~m} \mathrm{sec}^{-2}
$$

The final speed at which the ball reaches the surface of the sea:

$$
v=\sqrt{2 \gamma s}
$$

Therefore, the kinetic energy:

$$
\frac{m v^{2}}{2}=\frac{1}{2} \cdot \frac{d^{3} \pi}{6} \cdot \frac{\sigma}{g} \cdot 2 s \frac{\sigma-\sigma_{1}}{\sigma} g=\frac{d^{3} \pi}{6} \cdot s \cdot\left(\sigma-\sigma_{1}\right)=2588 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{sec}^{-2}
$$

The time of motion:

$$
t=\sqrt{\frac{2 s}{\gamma}}=4.89''
$$

(Geza Sasvari, Pécs.)

The problem was also solved by: Bayer B., Czank K., Faith F., Kerekes T., Krausz B., Krisztian Gy., Kürth A., Lukhaub Gy., Lupsa Gy., Scharff J., Szmodics K., Spitzer H., Weisz A.