A bullet fired into a fixed target loses half of its velocity after penetrating $3cm$. How much further will it penetrate before coming to rest assuming that it faces constant resistance to motion?

The correct option is B $1.0cm$
Let initial velocity of the bullet $=u$
After penetrating $3cm$ its velocity becomes $\frac{u}{2}$
From $v^2=u^2-2as$ (assuming constant deceleration $a$)
${\left(\frac{u}{2}\right)}^2=u^2-2a\left(3\right)$
$\Rightarrow 6a=\frac{3u^2}{4}\Rightarrow a=\frac{u^2}{8}$
Let it further penetrate through distance $x$ and stop at point $C$.
From B to C,
Final velocity$=0$, Intial velocity $=u/2$, $s=x,a=u^2/8$
So,
$v^2={\left(\frac{u}{2}\right)}^2-2as$
$\Rightarrow 0={\left(\frac{u}{2}\right)}^2-2\left(\frac{u^2}{8}\right).x$
$\Rightarrow x=1cm$