# A bullet fired into a plank loses half of its velocity after penetrating a thickness of $3 cm$. assuming that it has constant force of resistance, how much further does it penetrate before coming to rest

$(a)\;4\;cm\quad (b)\;3\; cm \quad (c)\;2\; cm \quad (d)\;1\;cm$

Let 'u' be initial velocity
$\bigg(\large\frac{u}{2}\bigg)^2$$-u^2=-2a \times 3 Since it loses half its velocity after 3 cm or a=\large\frac{u^2}{8} Let x be the further distance to which it penetrates Therefore Total distance =x'=3+x 0-u^2=-2 \large\frac{u^2}{8}$$x'$
$x'=4$
Therefore $x=4-3=1\; cm$
Hence d is the correct answer.
edited Mar 26, 2014