At time t = 0 , a disk of radius 1   m starts to roll without slipping on a horizontal plane with an…

Using second equation of motion,

The angle rotated by disc in $t=\sqrt{\pi } \mathrm{s}$ is $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$

$\Rightarrow \theta =\frac{1}{2}\times \frac{2}{3}{\left(\sqrt{\pi }\right)}^2$

$=\frac{\pi }{3} \mathrm{rad}$

Now using first equation of motion, the angular velocity of disc is $\omega ={\omega }_0+\alpha t$

$\Rightarrow \omega =0+\frac{2\sqrt{\pi }}{3}=\frac{2\sqrt{\pi }}{3} \mathrm{rad} {\mathrm{s}}^{-1}$

And, velocity of disk about centre of mass is $v_{\mathrm{cm}}=\omega R=\frac{2\sqrt{\pi }}{3}\times 1$

$=\frac{2\sqrt{\pi }}{3} \mathrm{m} {\mathrm{s}}^{-1}$

At the moment the stone detaches, the situation is shown in figure below.

$v=\sqrt{{\left(\omega R\right)}^2+v_{\mathrm{cm}}^2+2\left(\omega R\right)v_{\mathrm{cm}}\cos 120°}$

Putting the values, we get 

$v=\frac{2\sqrt{\pi }}{3} \mathrm{m} {\mathrm{s}}^{-1}$

And $\tan \theta =\frac{\omega R\sin 120°}{v_{\mathrm{cm}}+\omega R\cos 120°}$

$\Rightarrow \tan \theta =\sqrt{3}$

$\Rightarrow \theta =\frac{\pi }{3}\mathrm{rad}$

So, the maximum height reached by the stone is $H_{\max }=\frac{v^2{\sin }^2\theta }{2g}$

$=\frac{{\left(\frac{2\sqrt{\pi }}{3}\right)}^2\times {\sin }^{2}60°}{2\times 10}$

$=\frac{4\pi \times 3}{9\times 2\times 10\times 4}$

$=\frac{\pi }{60} \mathrm{m}$

So, height from ground will be

$R\left(1-\cos 60°\right)+\frac{\pi }{60}=\frac{1}{2}+\frac{x}{10}$

$\Rightarrow x=\frac{\pi }{6}=0.52$