A bullet is fired on a target with velocity ' \(\mathrm{V}\) '. Its velocity decreases from ' \(\mathrm{V}\) ' to ' \(\mathrm{V} / 2\) ' when it penetrates \(30 \mathrm{~cm}\) in a target. Through what thickness it will penetrate further in the target before coming to rest?

When the velocity of the bullet changes from \(\mathrm{V}\) to \(\frac{\mathrm{V}}{2}\) the distance travelled by the bullet is \(30 \mathrm{~cm}\).
Using \(3^{\text {rd }}\) equation of motion,
\(
\begin{aligned}
& \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{a} \\
& \left(\frac{\mathrm{V}}{2}\right)^2=\mathrm{V}^2+2 \mathrm{a}(30) \\
& \frac{\mathrm{V}^2}{4}=\mathrm{V}^2+60 \mathrm{a} \\
& \frac{-3 \mathrm{~V}^2}{4}=60 \mathrm{a} \\
& \mathrm{a}=\frac{-\mathrm{V}^2}{80}
\end{aligned}
\)
Further, when a bullet penetrates it comes to rest. So, the final velocity of the bullet becomes zero.
Using the relation,
\(
\begin{aligned}
& \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as} \\
& 0=\left(\frac{\mathrm{V}}{2}\right)^2+2\left(-\frac{\mathrm{V}^2}{80}\right) \mathrm{s} \\
& \frac{\mathrm{V}^2}{4}=\left(\frac{\mathrm{V}^2}{40}\right) \mathrm{s} \\
& \mathrm{s}=\frac{40}{4} \\
& \mathrm{~s}=10 \mathrm{~cm}
\end{aligned}
\)