Explanation

(c) Let the velocity of the bullet when it strikes the target is $v\ cm\ s^{- 1}.$
After penetrating 30 cm, velocity becomes half $ie,\frac{v}{2}.$
From equation $v^{2} = u^{2} + 2as$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \frac{v}{2} \right)^{2} = v^{2} + 2a \times 30$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - 60a = v^{2} - \frac{v^{2}}{4}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - 60a = \frac{3v^{2}}{4}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = - \frac{v^{2}}{80}cm\ s^{- 2}$
Let the bullet further penetrates $x$ cm before coming to rest, therefore
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v^{'2} = u^{'2} + 2\ as'$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 = \left( \frac{v}{2} \right)^{2} + 2\left( - \frac{v^{2}}{80} \right)\ x$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{v^{2}x}{40} = \frac{v^{2}}{4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 10\ cm$