Solution

From the geometry of the drawing, a trigonometric expression can be obtained by

$$\tan\varphi=\frac{l}{h}\tag{1}$$

$$\sin\varphi=\frac{l}{\sqrt{l^2+h^2}};\quad\cos\varphi=\frac{h}{\sqrt{l^2+h^2}}\tag{2}$$

Over the long period of time the $2m$ weight would go down infinitely ($h\to \infty$) From $(1)$, we obtain that as $h$ increases $\tan\varphi$ decreases

$$\tan\varphi \to 0;\quad\varphi \to 0$$

For the small angle $\varphi$ we could use an approximation to the second order of magnitude

$$\sin\varphi \approx\tan\varphi\approx\varphi;\quad\cos\varphi\approx 1$$

From the geometry of the figure, the length of the thread

$$\sqrt{l^2+h^2}=l+\Delta x$$

by the Pythagorean theorem, the change in the bottom of the thread connecting $2m$

$$\Delta x = \sqrt{l^2+h^2}-l$$

Similarly, from the geometry of the figure

Considering $\sin\varphi\approx\tan\varphi$ and $h\gg l$, we obtain

$$\sqrt{l^2+h^2}\approx h\Rightarrow \boxed{l=h-\Delta x}\tag{2}$$

$L_1,~L_2$ — Length of threads at the initial moment

Conservation of the mechanical energy $(dE_p+dE_k=0)$

Two weights with mass $m$ went up to $\Delta x$ and the $2m$ went down $h$, making change in potential energy

$$dE_p=-2mgh+2\cdot mg\Delta x$$

In the meantime, the velocities of $m$ and $2m$ bodies become $v_1$ and $v_2$, respectively, making total kinetic energy of the system

$$dE_k = 2\frac{mv_1^2}{2}+\frac{2m\cdot v_2^2}{2}$$

Substituing into equations of conservation of the mechanical energy

$$dE_p+dE_k=0\Leftrightarrow -2mgh+2\cdot mg\Delta x+2\frac{mv_1^2}{2}+\frac{2m\cdot v_2^2}{2}=0$$

$$v_1^2+v_2^2=2g\left(h-\Delta x\right)\tag{3}$$

By the Newton's second law for the $m$ weight

$$ma=T-mg$$

Similarly, for the $2m$ weight

$$2ma^*=2mg-2T\cos\varphi\approx2mg-2T\Leftrightarrow$$

$$\Leftrightarrow ma^*=mg-T$$

As they are connected by the same inextensible thread, their acceleration will be equal

$$\left|a\right|=\left|a^*\right|\Rightarrow v=v^*$$

Given that the system is being run from a stationary state $(v_0=v_0^*=0)$, let's substitute into $(3)$

$$2v^2=2g\left(h-\Delta x\right)$$

Taking into consideration $(2)$

$$v^2=gl\Rightarrow \boxed{v=\sqrt{gl}}$$