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+### Statement
+
+$2.3.16.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$2.3.16.$ A weight suspended on a thread of length $l$ was deflected by a distance $r$ from the equilibrium point and released. What is its maximum velocity?
+
+### Solution
+
+Let's consider the geometry of the system. Let the maximum deflection angle of the thread be $2\alpha$. The distance $r$ is the chord between the initial deflection point $D$ and the lowest equilibrium point $C$.
+The vertical height change $h$ (which corresponds to $BC$ in the diagram) can be expressed in two ways:
+$$h = l(1 - \cos 2\alpha) = r \sin\alpha$$
+
+Using the trigonometric identity $1 - \cos 2\alpha = 2 \sin^2\alpha$, we get:
+$$2l \sin^2\alpha = r \sin\alpha$$
+From this, we express the sine of half the deflection angle $\alpha$:
+$$\sin\alpha = \frac{r}{2l} \quad \text{(a)}$$
+
+From the law of conservation of energy, the kinetic energy $E_{\text{kin}}$ will be maximal at the lowest point $C$, where the potential energy is minimal. The maximum kinetic energy equals the initial potential energy:
+$$\frac{mv_{\text{max}}^2}{2} = mgh = mgr \sin\alpha$$
+Canceling the mass $m$:
+$$\frac{v_{\text{max}}^2}{2} = gr \sin\alpha$$
+
+Expressing the maximum velocity of the weight:
+$$v_{\text{max}} = \sqrt{2gr \sin\alpha} \quad \text{(b)}$$
+
+Substitute the angle expression from (a) into equation (b):
+$$v_{\text{max}} = \sqrt{2g \cdot r \frac{r}{2l}} = r\sqrt{\frac{g}{l}}$$
+
+#### Answer
+$$v_{\text{max}} = r\sqrt{\frac{g}{l}}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]