Spherical ball of radius $r$ initially at rest on a rough horizontal surface is hit horizontally at a point at a distance $x=\frac{r}{2}$  above the centre line. Due to this sharp impulse the centre of the ball acquires a velocity  $V_c$. After some time, the ball will start pure rolling with a velocity equal to

Initially ball will gain both linear and angular velocity. Linear impulse:  $I_0=m{v}_C$
Angular impulse :  $I_{0}x=I{\omega }_0\Rightarrow m{v}_{C}x=I{\omega }_0$
   
$I$  is moment of inertia about C.
Apply conservation of angular momentum about lowest point.
$I{\omega }_0+m{v}_{C}r=I\omega +mvr$ 
Where  $\omega =\frac{v}{r}$ at the time of pure rolling
$$\begin{gathered} \Rightarrow m{v}_{C}x+m{v}_{C}r=\frac{2}{5}m{r}^2\frac{v}{r}+mvr \\ \Rightarrow v_C(x+r)=\frac{7}{5}rv \\ \Rightarrow v=\frac{7}{5}{v}_{{ }_C}\left[\frac{x+r}{r}\right] \end{gathered}$$