\(\tan \theta=\frac{v_y}{v_x}\)

Also, \(t_1+t_2=\frac{2 u \sin \theta}{g}\)

\(4=\frac{2 \times 40 \times \sin \theta}{10}\)

\(\sin \theta=\frac{1}{2} \Rightarrow \theta=30^{\circ}\)

So, \(\tan \theta=\tan 30^{\circ} \Rightarrow \frac{1}{\sqrt{3}}\)

\(\theta=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)