At time t = 0 , a disk of radius 1m starts to roll without slipping...

Using second equation of motion,

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

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

$= \dfrac{\pi}{3}rad$

Now using first equation of motion, the angular velocity of disc is $\omega = {\omega}_{0} + \alphat$

$⇒ \omega = 0 + \dfrac{2\sqrt{\pi}}{3} = \dfrac{2\sqrt{\pi}}{3}rad{s}^{- 1}$

And, velocity of disk about centre of mass is ${v}_{cm} = \omegaR = \dfrac{2\sqrt{\pi}}{3} \times 1$

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

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

$v = \sqrt{{\left(\omegaR\right)}^{2} + {v}_{cm}^{2} + 2\left(\omegaR\right){v}_{cm}\cos 120 °}$

Putting the values, we get

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

And $tan\theta = \dfrac{\omega Rsin120 °}{{v}_{cm} + \omega Rcos120 °}$

$⇒ tan\theta = \sqrt{3}$

$⇒ \theta = \dfrac{\pi}{3}rad$

So, the maximum height reached by the stone is {H}_\max = \dfrac{{v}^{2}\sin^{2}\theta}{2g}

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

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

$= \dfrac{\pi}{60}m$

So, height from ground will be

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

$⇒ x = \dfrac{\pi}{6} = 0 . 52$