A circular track of radius 1 m is as shown in the figure. A small block is moving on horizontal part of the track with speed $v_0 = \sqrt{44} m/s$. The value of sin 𝜃 at which it leaves the track is x/5. Find the value of ‘x’
 

At the time when the block leaves the surface normal reaction between the block and the track is zero

$\Rightarrow \frac{{\mathrm{mv}}^2}{\mathrm{R}}=\mathrm{mgsin}\mathrm{\theta }+\mathrm{N}(\mathrm{N}=0)\Rightarrow {\mathrm{v}}^2=\mathrm{Rgsin}\mathrm{\theta }$

From conservation of mechanical energy, we can write

$\begin{array}{l}{\mathrm{v}}^2={\mathrm{u}}^2-2\mathrm{Rg}(1+\sin \mathrm{\theta })\Rightarrow \mathrm{Rgsin}\mathrm{\theta }={\mathrm{u}}^2-2\mathrm{Rg}-2\mathrm{Rgsin}\mathrm{\theta }\\ \Rightarrow \sin \mathrm{\theta }=\frac{{\mathrm{u}}^2-2\mathrm{Rg}}{3\mathrm{Rg}}=\frac{44-2\times 1\times 10}{3\times 1\times 10}=\frac{24}{30}=\frac{4}{5}\end{array}$