Geometry of P. With $C$ at the top, the centre $O$ is at distance $R=0.2$ m below $C$. Since $OP$ makes $45^{\circ }$ with the vertical, $P$ lies at horizontal distance $R/\sqrt{2}$ from the vertical through $C$ and at vertical distance $R+R/\sqrt{2}$ below $C$ (using the geometry shown). The perpendicular distance of the incoming horizontal velocity (along $-\hat{i}$) from $C$ equals the vertical drop of $P$ below $C$, which is $R+R/\sqrt{2}=R\left(1+\frac{1}{\sqrt{2}}\right)$. The perpendicular distance for the outgoing velocity (along $-\hat{j}$) equals the horizontal distance of $P$ from $C$, which is $R/\sqrt{2}$.

Angular momentum about $C$ (conserved during the instantaneous collision).

Moment of inertia of the disk about $C$ (using parallel axis): $I_C=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2=\frac{3}{2}(1)(0.2{)}^2=0.06$ kg m${}^2$.

$m{v}_{i}R\left(1+\frac{1}{\sqrt{2}}\right)=I_C\omega +m{v}_f\frac{R}{\sqrt{2}}$

$(0.02)(100)(0.2)(1+0.707)=0.06\omega +(0.02)(90)(0.2)(0.707)$

$0.6828=0.06\omega +0.2546\Rightarrow \omega \approx 7.14$ rad/s.

Maximum rise of $O$ from energy conservation. After collision the disk rotates about $C$, and gravity acts on its centre $O$:

$\frac{1}{2}{I}_C{\omega }^2=MgR(1-\cos {\theta }_{\max })$

$\frac{1}{2}(0.06)(7.14{)}^2=(1)(10)(0.2)(1-\cos {\theta }_{\max })$

$1.529=2(1-\cos {\theta }_{\max })$, but the relevant quantity is the vertical rise $\Delta h=R(1-\cos {\theta }_{\max })$:

$\Delta h=\frac{I_C{\omega }^2}{2Mg}=\frac{(0.06)(7.14{)}^2}{2(1)(10)}\approx 0.153$ m $\approx 0.15$ m.