c
(c) If suppose bob rises up to a height h as shown then after releasing potential energy at extreme position becomes kinetic energy of mean position

\( \Rightarrow mgh = \frac{1}{2}mv_{\max }^2\)

\( \Rightarrow {v_{\max }} = \sqrt {2gh} \)

Also, from figure \(\cos \theta = \frac{{l - h}}{l}\)

\( \Rightarrow h = l(1 - \cos \theta )\)

So, \({v_{\max }} = \sqrt {2gl(1 - \cos \theta )} \)