1. A boat starts moving downstream along the river at 10 o'clock. At the same moment, a ball is thrown from it. After 15 minutes, the boat turns back to catch the ball. At what time will the boat meet the ball?
Solution. Let the speed of the river be $v$, and the speed of the boat in still water be $v_{1}$. If the boat started from point $A$ and after $15 \mathrm{~min}$ arrived at point $B$ (make a diagram), it traveled the distance $\overline{A B}=\frac{1}{4}\left(v+v_{1}\right) \mathrm{km}$. In the same time, the ball moved from $A$ to $C$ and traveled the distance $\overline{A C}=\frac{1}{4} v$.

From point $B$, the boat turns back. Suppose it met the ball at point $E$ after $x$ hours. Then $\overline{B E}=x\left(v_{1}-v\right), \overline{C E}=x v$, so from

$$
\overline{A C}+\overline{C E}+\overline{E B}=\overline{A B}
$$

and the previous equations, we get $x=\frac{1}{4}$ hour $=15$ minutes. Therefore, the boat will meet the ball at 10:30.