Explanation

(c) [image:5_image292.emf]
Let bullet is fired with velocity $v_{B}$ at point $A$ and its velocity becomes half when it travels a distance $s$ and reaches at point $B$. When it reaches at point $C$, it comes to rest and travels a distance $x$
From $A$ to $B$, using, $v^{2} - u^{2} = 2as$
$\Rightarrow \left( \frac{v_{B}}{2} \right)^{2} - v_{B}^{2} = 2as \Rightarrow \frac{v_{B}^{2}}{4} - v_{B}^{2} = 2as$
$\Rightarrow \frac{- 3v_{B}^{2}}{4} = 2as \Rightarrow a = \frac{- 3v_{B}^{2}}{8s}$
$\therefore$ From$\ B\ to\ C\ ,$ using$\ v^{2} - u^{2} = 2as$
$0^{2} - \left( \frac{v_{B}}{2} \right)^{2} = 2as\ \Rightarrow \frac{- v_{B}^{2}}{4} = 2\left( \frac{- 3v_{B}^{2}}{8s} \right)x$
$\Rightarrow x = \frac{8s}{4 \times 6} = \frac{8 \times 30}{24} = 10\ cm$