(b, d, f) Hint: Apply the concept of conservation of energy.

Step 1: Find the velocity of the bullet just before hitting the target.

Consider the adjacent diagram for the given situation in the question.

(b) Conserving energy between "O" and "A"

$\begin{array}{l}{\mathrm{U}}_{\mathrm{i}}+{\mathrm{K}}_{\mathrm{i}}={\mathrm{U}}_{\mathrm{t}}+{\mathrm{K}}_{\mathrm{t}}\\ \Rightarrow 0+\frac{1}{2}{\mathrm{mv}}^2=\mathrm{mgh}+\frac{1}{2}{\mathrm{mv}}^{\mathrm{\prime }}\\ \Rightarrow \frac{({\mathrm{v}}^{\mathrm{\prime }}{)}^2}{2}=\frac{{\mathrm{v}}^2}{2}=-\mathrm{gh}\\ \Rightarrow ({\mathrm{v}}^{\mathrm{\prime }}{)}^2={\mathrm{v}}^2-2\mathrm{gh}\Rightarrow {\mathrm{v}}^{\mathrm{\prime }}=\sqrt{{\mathrm{v}}^2-2\mathrm{gh}} ...\left(\mathrm{i}\right)\end{array}$

where v' is of the bullet just before hitting the target.

Step 2: Find the velocity of the bullet just after hitting the target.

Let after emerging from the target is 'v' then,

By question,

$\begin{array}{l} =\frac{1}{2}({\mathrm{mv}}^{\prime \prime }{)}^2=\frac{1}{2}[\frac{1}{2}\mathrm{m}\left({\mathrm{v}}^{\mathrm{\prime }}{)}^2\right]\\ \frac{1}{2}\mathrm{m}({\mathrm{v}}^{\prime \prime }{)}^2=\frac{1}{4}\mathrm{m}({\mathrm{v}}^{\mathrm{\prime }}{)}^2=\frac{1}{4}\mathrm{m}[{\mathrm{v}}^2-2\mathrm{gh}]\\ \Rightarrow ({\mathrm{v}}^{\prime \prime }{)}^2=\frac{{\mathrm{v}}^2-2\mathrm{gh}}{2}=\frac{{\mathrm{v}}^2}{2}-\mathrm{gh}\\ \Rightarrow {\mathrm{v}}^{\prime \prime }=\sqrt{\frac{{\mathrm{v}}^2}{2}-\mathrm{gh}} ...\left(\mathrm{ii}\right)\end{array}$

Step 3: Find the relation between the two velocities.

From Eqs. (I) and (ii)

$\begin{array}{l} \frac{{\mathrm{v}}^{\mathrm{\prime }}}{{\mathrm{v}}^{\prime \prime }}=\frac{\sqrt{{\mathrm{v}}^2-2\mathrm{gh}}}{\frac{\sqrt{{\mathrm{v}}^2-2\mathrm{gh}}}{\sqrt{2}}}=\sqrt{2}\\ \Rightarrow {\mathrm{v}}^{\prime \prime }=\frac{{\mathrm{v}}^{\mathrm{\prime }}}{\sqrt{2}}={\mathrm{v}}^2\left(\frac{{\mathrm{v}}^{\mathrm{\prime }}}{2}\right)\\ \Rightarrow \frac{{\mathrm{v}}^{\prime \prime }}{\frac{{\mathrm{v}}^{\mathrm{\prime }}}{2}}=\sqrt{2}=1.414>1\\ \Rightarrow {\mathrm{v}}^{\prime \prime }>\frac{{\mathrm{v}}^{\mathrm{\prime }}}{2}\end{array}$

Hence. after emerging from the target velocity of the bullet (v" ) is more than half of earlier velocity v' (velocity emerging into the target).

(d) As the velocity of the bullet changes to v' which is less than 'v' hence. path. followed will change and the bullet reaches point B instead of ''A', as shown in the figure.

(f) As the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target. Therefore, their internal energy increases.