Explanation

(a) Velocity of bullet at highest point of its trajectory = 50 $\cos\theta$ in horizontal direction.
As bullet of mass $m$ collides with pendulum bob of mass 3$m$ and two stick together, their common velocity
$v^{'} = \frac{m_{1}50\cos\theta}{m + 3n} = \frac{25}{2}\cos{\theta ms^{- 1}}$
As now under this velocity $v'$ pendulum bob goes up to an angel 120${^\circ}$, hence
$\frac{v^{'2}}{2g} = h = l\left( 1 - \cos{120{^\circ}} \right) = \frac{10}{3}\left\lbrack 1 - \left( f - \frac{1}{2} \right) \right\rbrack = 5$
$\Rightarrow \ \ v^{'2} = 2 \times 10 \times 5 = 100$ or $v^{'} = 10$
Comparing two answer of $v'$, we get
$\frac{25}{2}\cos{\theta = 10 \Rightarrow \cos{\theta = \frac{4}{5}}}$ or$\ \theta = \cos^{- 1}\left( \frac{4}{5} \right)$