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When, we throw the particle with minimum required velocity, it touches the cylinder at two points. Let us say the line joining center of cylinder and particle makes an angle $\phi$ with horizontal, and at this time speed is $v$ . Clearly, height scaled is $h = R(1 + \sin \phi)$ .

Using conservation of energy,

$v^2 = u^2 - 2gh = u^2 -2gR(1+\sin \phi)$

Now, the range of particle if it is thrown from this point with speed $v$ is :

$\displaystyle x = \frac{v^2 \sin 2\big(\frac{\pi}{2} - \phi\big)}{g} = \frac{(u^2 - 2gR(1+\sin \phi))\sin 2 \phi}{g}$

Now, this should be same as $2R \cos \phi$ . as the distance between two points where particle touches cylinder is $2R \cos \phi$

Hence, $\displaystyle \frac{(u^2 - 2gR(1+\sin \phi))\sin 2 \phi}{g} = 2R \cos \phi$

$\displaystyle \Rightarrow u^2 = gR(2 + 2 \sin \phi + \frac{1}{\sin \phi}) \geq 2gR(1+ \sqrt{2})$ ( AM-GM)

Thus , $u \geq 9.728 m/s$