Physics - Kinematics Question with Solution | TestHub

The bomb's initial horizontal velocity is $v$.

The time to fall H is $t=\sqrt{2H/g}$.

The horizontal distance covered is $x=vt=v\sqrt{2H/g}$.

The angle $\theta$ is given by $\tan \theta =x/H=(v\sqrt{2H/g})/H=v\sqrt{2/gH}$.

Thus, $\theta ={\tan }^{-1}(v\sqrt{2/gH})$.