Explanation

(c) For projectile $A$
Maximum height, $H_{A} = \frac{u_{A}^{2}\sin^{2}{45{^\circ}}}{2g}$
For projectile $B$
Maximum height $H_{B} = \frac{u_{B}^{2}\sin^{2}\theta}{2g}$
As per equation
$H_{A} = H_{B}$
$\frac{u_{A}^{2}\sin^{2}{45{^\circ}}}{2g} = \frac{u_{B}^{2}\sin^{2}\theta}{2g}$
$\Rightarrow \frac{\sin^{2}\theta}{\sin^{2}{45{^\circ}}} = \frac{u_{A}^{2}}{u_{B}^{2}}$
$\Rightarrow \sin^{2}\theta = \left( \frac{u_{A}}{u_{B}} \right)^{2}\sin^{2}{45{^\circ}}$
$\Rightarrow \sin^{2}\theta = \left( \frac{1}{\sqrt{2}} \right)^{2}\left( \frac{1}{\sqrt{2}} \right)^{2}\ \ \left\lbrack \because\frac{u_{A}}{u_{B}} = \frac{1}{\sqrt{2}}\ (Given) \right\rbrack$
$\Rightarrow \sin\theta = \frac{1}{2} \Rightarrow \theta\sin^{- 1}\left( \frac{1}{2} \right) = 30{^\circ}$