A shell bursts on contact with ground and pieces from it fly in all directions with a velocity equal to 60 m/s. It is known that a man who is 180 m away is in danger for a time interval of  $\sqrt{x}$ sec, find x ?(Neglect height of man).

$$\begin{gathered} u=60m/s R=180m/s \\ R=\frac{u^2\sin 2\theta }{g}\Rightarrow Sin2\theta =\frac{180\times 10}{60\times 60}=\frac{1}{2} \\ \Rightarrow 2\theta ={30}^0\Rightarrow \theta ={15}^0 \end{gathered}$$

Two directions of projections have same range  ${\theta }^0\&{90}^0-{\theta }^0$
Let ,  $T_1\&T_2$ are time of flights in 2 cases for  
${\theta }_1={15}^0$  and   ${\theta }_2=90-15={75}^0$
  $T_1=\frac{2u\sin {\theta }_1}{g} T_2=\frac{2u\sin {\theta }_2}{g}$
Man  is in danger for a time 
$$\begin{gathered} T_2-T_1=\frac{2u}{g}(\sin {75}^0-\sin {15}^0) \\ =\frac{2u}{g}\left(2\cos {45}^0\sin {30}^0\right) \\ =\frac{2\times 60}{10}\times 2\times \frac{1}{\sqrt{2}}\times \frac{1}{2} \\ =6\sqrt{2}sec \\ =\sqrt{72} sec \end{gathered}$$