In figure, lower pulley is free to move in vertical direction only. Block A is given a uniform velocity u as shown, what is velocity of block B as a function of angle $\theta$.

Let in a small time dt, displacement of A is dx and displacement of B is dy. Increase in length l₁= dx.

Increase in length $l_2=dx\sin \theta -dy\cos \theta$

But net increase in length should be zero, so

$dx+dx\sin \theta -dy\cos \theta =0$

$\begin{array}{l}\Rightarrow dy=\frac{(1+\sin \theta )dx}{\cos \theta }\\ \Rightarrow \frac{dy}{dt}=\left(\frac{1+\sin \theta }{\cos \theta }\right)\frac{dx}{dt}\Rightarrow v=\left(\frac{1+\sin \theta }{\cos \theta }\right)u\end{array}$