Explanation

(b) The body covers a horizontal distance AB during its flight. This horizontal range is given by
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$R = \frac{u^{2}\sin{2\theta}}{g}\ \ \ \ \ \ \ldots.(i)$
$For\ maximum\ horizontal\ range,\sin{2\theta = 1}$
$\therefore\ R_{\max} = \frac{u^{2}}{g}\ \ \ \ldots(ii)$
$Given,\ R = 3km,\ \ \theta = 30{^\circ}$
$\therefore\ \ \ From\ Eq.(i)$
$\frac{u^{2}}{g} = \frac{R}{\sin{2\theta\ }} = \frac{3}{\sin{60{^\circ}}} = \frac{3 \times 2}{\sqrt{3}} = \sqrt{3} \times 2$
$\frac{u^{2}}{g} = 3.464m\ \ \ or\ R_{\max} = 3.46cm$
Hence, maximum range with velocity of projection $u$ cannot be more than 3.464 m, Hence, it is not possible to hit a target 5 km away.