Let us assume that the ball is coming from right. One can solve it by conserving angular momentum about point of contact of sphere and ground(let it be point O).

After colliding with the wall the velocity of the CoM of the sphere will be $\frac{V}{2}$ .

However the angular velocity of the sphere will be $\frac{V}{r}$ just after the collision. Because during collision no impulsive torque acts on the sphere. So angular velocity of sphere before collision is equal to angular velocity of sphere after collision. After that as direction of frictional force changes so torque due to friction reduce the angular velocity of the sphere.

Now

Initial angular momentum of sphere about point O= $\frac { mvr }{ 2 } -I{ \omega }_{ i }$

Let the speed of CoM be $V_{f}$ when the sphere starts rolling. So $\omega_{f}=\frac{V_{f}}{r}$

So final angular momentum of sphere about point O= $m{ V }_{ f }r+I{ \omega }_{ f }$

So

$\frac { mvr }{ 2 } -I{ \omega }_{ i }=m{ V }_{ f }r+I{ \omega }_{ f }$ (I have taken here clockwise direction as -ve here)

By putting $I=\frac { 2m{ r }^{ 2 } }{ 5 }$ , ${ \omega }_{ i }=\frac { V }{ r }$ and ${ \omega }_{ f }=\frac { { V }_{ f } }{ r }$ we get

${ V }_{ f }=\frac { V }{ 14 }$

Now initial velocity of the CoM of sphere $u$ = $\frac{V}{2}$

final velocity of the CoM of sphere is $V_{f}$

and acceleration of the CoM of sphere = $-\mu g$

As $v_{f}=u+at$

So $t=\frac { 3V }{ 7\mu g }$

:)

Let us assume that before collision the body is travelling to the right.

After collision it moves to the left and has velocity $\frac{v}{2}$ and same angular velocity (Using newton's experimental law and conservation of angular momentum about the impact point).

We find that the contact point with the ground has velocity to the left. Therefore the frictional force $\mu mg$ acts towards right.

Now let ${V}_{CM}$ be the velocity of sphere and $\frac{{V}_{CM}}{R}$ be its angular velocity when it begins pure rolling.

${ v }_{ CM }=\frac { v }{ 2 } -\mu gt$

$\frac { { v }_{ CM } }{ r } =\frac { -v }{ r } +\frac { 5\mu mgr }{ 2m{ r }^{ 2 } } t$

Solving the two equation we obtain,

$t=\frac { 3v }{ 7\mu g }$

Question --> OVERRATED