A bullet is fired into a fixed target looses one third of its velocity after travelling \(4 cm\). It penetrates further …

\begin{equation}
\begin{aligned}
&\begin{aligned}
& v^2-u^2=2 a S \\
& \left(\frac{2 u}{3}\right)^2=u^2+2(-a)\left(4 \times 10^{-2}\right) \\
& \Rightarrow \frac{4 u^2}{9}=u^2-2 a\left(4 \times 10^{-2}\right) \\
& \Rightarrow-\frac{5 u^2}{9}=-2 a\left(4 \times 10^{-2}\right) \\
& \Rightarrow 0=\left(\frac{2 u}{3}\right)^2+2(-a)(x) \\
& \Rightarrow-\frac{4 u^2}{9}=-2 a x
\end{aligned}\\
&\text { Dividing equation (1) by (2), we get }\\
&\begin{aligned}
& \frac{5}{4}=\frac{4 \times 10^{-2}}{x} \Rightarrow x=\frac{16}{5} \times 10^{-2} \\
& \Rightarrow x=3.2 \times 10^{-2} \mathrm{~m}=32 \times 10^{-3} \mathrm{~m}
\end{aligned}
\end{aligned}
\end{equation}