A shell bursts on contact with the ground and pieces from it fly in...

Given, $u = 60m{s}^{- 1}, R = 180m$

$∴ R = \dfrac{{u}^{2}}{g}\sin 2\theta$

$180 = \dfrac{{\left(60\right)}^{2}}{10}\sin 2\theta$

$⇒ \sin 2\theta = \dfrac{1}{2}$

$⇒ \theta = {15}^{o}$

As there are always two directions of projections $\theta$ and \left({90}^{o} - \theta\right)$for the given range, hence two directions of projection, i.e.${15}^{o}$and${90}^{o} - {15}^{o},$i.e.${15}^{o}$and${75}^{o} .$Let${T}_{1}$and${T}_{2}$be the times of flight in the two cases, where<br /><br />${T}_{1} = \dfrac{2usin {15}^{o}}{g}$and${T}_{2} = \dfrac{2usin {75}^{o}}{g}$<br /><br />The man is in danger for a time<br /><br />${T}_{2} - {T}_{1} = \dfrac{2u}{g}\left(sin {75}^{o} - sin {15}^{o}\right)$<br /><br />$= \dfrac{2u}{g}\left(2cos {45}^{o}sin {30}^{o}\right)$<br /><br />\left[usingidentity, sin C - sin D = 2cos \dfrac{C + D}{2}sin \dfrac{C - D}{2}\right]$

$= \dfrac{2 \times 60}{10} \times 2 \times \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2}$

$= \sqrt{72}s$