A container with vertical walls stands on a horizontal plane, in which the water stands at a height of $a$. At which point on the side wall should a hole be drilled if we want to achieve that the jet of water emerging from it hits the horizontal plane at the point farthest from the base of the container?
Let the hole be at a height $x$ measured from the bottom of the vessel, then the speed of the outflowing water is

$$
V=\sqrt{2 g(a-x)}
$$

and if the water jet hits the horizontal plane at a distance $y$, then

$$
y=t \cdot \sqrt{2 g(a-x)}
$$

where $t$ represents the time of fall; this is the same as if it fell freely from a height of $x$, so

$$
\begin{gathered}
t=\sqrt{\frac{2 x}{g}} \\
y=2 \sqrt{a x-x^{2}}
\end{gathered}
$$

This function has a maximum when

$$
x=\frac{a}{2}
$$

in which case

$$
y_{\max }=a
$$

(Jánosy Gyula, Budapest.)

The problem was also solved by: Bánó L., Csada I., Fodor H., Merse P., Pichler S., Rosenberg J., Schuster Gy., Székely J., Szilas O.