For the arrangement of mirror and object shown in figure, the velocity of image w.r.t ground is:

The correct option is B $-5(1+\sqrt{3})\hat{i}+5\hat{j}\text{m/s}$
We know that for fixed plane mirror,
$\left(1\right)$ Perpendicular to the mirror ($x$ - axis), $v_o=-v_i$
$\left(2\right)$ Parallel to the mirror ($y$ - axis), $v_o=v_i$
The normal direction for plane mirror is $x$ - axis
$\Rightarrow (v_{o,m}{)}_x=-(v_{i,m}{)}_x$
$(v_o{)}_x-(v_m{)}_x=-\left[\right(v_i{)}_x-\left(v_m{)}_x\right]$
$\Rightarrow 10\cos {30}^{\circ }\hat{i}-(-5\cos {60}^{\circ })\hat{i}=-\left[\right(v_i{)}_x-(-5\cos {60}^{\circ }\hat{i})]$
$\Rightarrow 5\sqrt{3}\hat{i}+\frac{5}{2}\hat{i}=-(v_i{)}_x-\frac{5}{2}\hat{i}$
$\therefore (v_i{)}_x=-(5+5\sqrt{3})\hat{i}\text{m/s}$

Along $y$ - direction.
$(v_0{)}_y=(v_i{)}_y=10\sin {30}^{\circ }\hat{j}$
$\therefore (v_i{)}_y=5\hat{j}\text{m/s}$
$\Rightarrow {\overrightarrow{v}}_i=-5(1+\sqrt{3})\hat{i}+5\hat{j}\text{m/s}$