$$a = \frac{{{{v''}_0} - {{v'}_0}}}{{\Delta t}}$$
On substituting
$${{v'}_0} = 2\,m/s,{{v''}_0} = 12\,m/s$$      and $$\Delta t = 10\,\sec ,$$
we get, $$a = 1\,m/{s^2}.$$

$$\eqalign{ & {t_1} = \frac{{2u\sin \theta }}{g}\,{\text{and}} \cr & {t_2} = \frac{{2u\sin \left( {90 - \theta } \right)}}{g} \cr & = \frac{{2u\cos \theta }}{g} \cr & \therefore {t_1}{t_2} = \frac{{4{u^2}\cos \theta \sin \theta }}{{{g^2}}} = \frac{2}{g}\left[ {\frac{{{u^2}\sin \theta }}{g}} \right] \cr & = \frac{2}{g}R, \cr} $$
where $$R$$ is the range. Hence $${t_1}{t_2} \propto R$$

Let $$s$$ be the distance travelled by the vehicle before it stops.
Final velocity $$\upsilon = 0,$$  initial velocity = $$u$$
Using equation of motion $${v^2} - {u^2} = 2aS$$
$${0^2} - {u^2} = 2aS$$
Stopping distance, $$S = - \frac{{{u^2}}}{{2a}}$$

From given equation,
$$\eqalign{ & \vec v = k\left( {y\hat i + x\hat j} \right) = ky\hat i + kx\hat j = {V_x}\hat i + {V_y}\hat j \cr & \frac{{dx}}{{dt}} = ky\,\,{\text{and}}\,\,\frac{{dy}}{{dt}} = kx \cr & {\text{Now}},\,\,\frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{x}{y} = \frac{{dy}}{{dx}}\,\,\, \Rightarrow ydy = xdx \cr} $$
Integrating both sides we get $${y^2} = {x^2} +$$   constant

For first part of penetration, by equation of motion
$${\left( {\frac{u}{2}} \right)^2} - {\left( u \right)^2} = 2aS\,\,{\text{or}}\,\,a = - \frac{{3{u^2}}}{{8S}}\,......\left( {\text{i}} \right)$$
For latter part of penetration
$$\eqalign{ & {\left( 0 \right)^2} - {\left( {\frac{u}{2}} \right)^2} = 2aS',\,S' = - \frac{{{u^2}}}{{8a}} \cr & S' = - \frac{{{u^2}}}{8}\left( {\frac{{8S}}{{ - 3{u^2}}}} \right)\,\,\left( {{\text{Using}}\,\left( {\text{i}} \right)} \right) \cr & S' = \frac{S}{3}\,\,{\text{or}}\,\,S' = \frac{{40}}{3}cm \cr} $$

Velocity when the engine is switched off
$$\eqalign{ & v = 19.6 \times 5 = 98\,m{s^{ - 1}} \cr & {h_{\max }} = {h_1} + {h_2} \cr & {\text{where}}\,{h_1} = \frac{1}{2}a{t^2}\,\& \,{h_2} = \frac{{{v^2}}}{{2a}} \cr & {h_{\max }} = \frac{1}{2} \times 19.6 \times 5 \times 5 + \frac{{98 \times 98}}{{2 \times 9.8}} = 735\,m \cr} $$

$$\eqalign{ & {\text{Given,}}\,g = 10\,m{s^{ - 2}}\,{\text{and}}\,h = 20\,m \cr & {\text{We}}\,{\text{have}}\,v = \sqrt {2gh} = \sqrt {2 \times 10 \times 20} \cr & = \sqrt {400} = 20\,m/s \cr} $$

$$\eqalign{ & {a_r} = {\omega ^2}R \cr & {a_r} = {\left( {2\pi 2} \right)^2}R = 4{\pi ^2}{2^2}R \cr & = 4{\pi ^2}{\left( {\frac{{22}}{{44}}} \right)^4}\left( 1 \right)\,\,\left[ {\because v = \frac{{22}}{{44}}} \right] \cr & {a_t} = \frac{{dv}}{{dt}} = 0 \cr} $$
$${a_{net}} = {a_r} = {\pi ^2}m{s^{ - 2}}$$    and direction along the radius towards the centre.

This situation is plotted on $$\left( {v - t} \right)$$  graph. In $$\left( {v - t} \right)$$  graph, $$OA$$  represents the accelerated part and $$AB$$  represents the decelerated part.
Let $${t_1}$$ and $${t_2}$$ be the times for part $$OA$$  and $$AB$$  respectively.
At point $$A$$ velocity is maximum and let it be $${v_{\max }}.$$
$$\eqalign{ & \therefore {v_{\max }} = \alpha {t_1} = \beta {t_2} \cr & {\text{But}}\,\,t = {t_1} + {t_2} = \frac{{{v_{\max }}}}{\alpha } + \frac{{{v_{\max }}}}{\beta } \cr & = {v_{\max }}\left( {\frac{1}{\alpha } + \frac{1}{\beta }} \right) \cr & = {v_{\max }}\left( {\frac{{\alpha + \beta }}{{\alpha \beta }}} \right) \cr & {\text{or}}\,\,{v_{\max }} = t\left( {\frac{{\alpha \beta }}{{\alpha + \beta }}} \right) \cr} $$
Alternative
This problem can also be solved by checking the dimensions on both sides. On checking the dimensions we note that the dimensions of option (D) match with that of velocity.

Average speed $$ = \frac{{{\text{Total distance travelled}}}}{{{\text{Total time taken}}}}$$
$$\eqalign{ & = \frac{x}{{\frac{{\frac{{2x}}{5}}}{{{v_1}}} + \frac{{\frac{{3x}}{5}}}{{{v_2}}}}} \cr & = \frac{{5{v_1}{v_2}}}{{3{v_1} + 2{v_2}}} \cr} $$