At the malomgátak, sloping channels are commonly used for the drainage of excess water, the cross-section of which is a right-angled parallelogram. Water arrives at the upper end of such a channel at a speed of $1.4 \frac{\mathrm{m}}{\text { sec }}$ and flows over it at a thickness of $0.4 \mathrm{~m}$. The channel is $10 \mathrm{~m}$ long, has a uniform width of $3.5 \mathrm{~m}$, and an inclination angle of $12^{\circ}$. What is the thickness of the water flow at the lower part of the channel? How wide would this channel need to be built so that the water flows at the same thickness at the lower part as at the upper part?
The amount of water flowing into the channel

$$
K=0.4 \cdot 3.5 \cdot 1.4 \frac{\mathrm{m}^{3}}{\mathrm{sec}}
$$

The velocity of the outflow at the bottom

$$
v=\left(1.4+\sqrt{2 \cdot 9.81 \cdot \sin 12^{\circ} \cdot 10}\right) \frac{\mathrm{m}}{\sec }
$$

Thus, the thickness of the outflow

$$
x=\frac{K}{v \cdot 3.4} \mathrm{~m}=0.07 \mathrm{~m}
$$

In the second case, the width of the channel

$$
y=\frac{K}{v \cdot 0.4} \mathrm{~m}=0.63 \mathrm{~m}
$$

(Oszkár Szilas, Budapest.)

The problem was solved by: L. Bánó, I. Csada, P. Ehrenstein, E. Freund, A. Haar, Gy. Jánosy, I. Koritsánszky, T. Pözel, J. Székely.