d
As \(P\) and \(Q\) fall down, the length l decreases at the rate of \(U \,m/s.\)

From the figure, \({l^2} = {b^2} + {y^2}\)

Differentiating with respect to time

\(2l \times \frac{{dl}}{{dt}} = 2b \times \frac{{db}}{{dt}} + 2y \times \frac{{dy}}{{dt}}\)      \(\left( {{\rm{As \,\,}}\frac{{db}}{{dt}} = 0,\frac{{dl}}{{dt}} = U} \right)\)

\(⇒\) \(\frac{{dy}}{{dt}} = \left( {\frac{l}{y}} \right) \times \frac{{dl}}{{dt}}\) \( \Rightarrow \frac{{dy}}{{dt}} = \left( {\frac{1}{{\cos \theta }}} \right) \times U = \frac{U}{{\cos \theta }}\)