Rotation 1

Let the impulse imparted by the cue be equal to $F \,dt$ .

Using impulse theorem, we have

$F \,dt = mv - m (0) = mv \qquad \cdots (1)$

Using angular impulse theorem about the point of contact of shell with the ground, we have

$F \,dt (H+R) = I \omega - I (0) = \left( I_{CM} + mR^2 \right) \omega = \left( \dfrac 23 mR^2 + mR^2 \right) \omega = \dfrac 53 mR^2 \omega \qquad \cdots (2)$

where $I$ is the moment of inertia of the shell about the contact point of shell and the ground.

Dividing $(2)$ by $(1)$ we have

$H+R = \dfrac{5R^2 \omega}{3v}$

For pure rolling, we must have $v = R \omega$ . Thus

$H+R = \dfrac{5R^2 \omega}{3v} = \dfrac{5R}{3} \implies H = \dfrac{2R}{3} = \dfrac{2 (6)}{3} = \boxed{4 \text{ cm}}$