A plane is inclined at an angle \(\alpha  = 30^o\) with respect to the horizontal. A particle is projected with a speed \(u = 2\,ms^{-1},\) from the base of the plant, making an angle \(\theta = 15^o\) with respect to the plane as shown in the figure. The distance from the base at which the particle hits the plane is close to  ........ \(cm\) (Take \(g = 10\,ms^2\) )

\(\begin{array}{l}
T = \frac{{2\,u\,\sin \theta }}{{g\,\cos \alpha }}\\
R\, = u\,\cos \theta \,T - \frac{1}{2}\,g\,\sin \,\alpha \,{T^2}\\
\,\,\,\,\, = \,\frac{{u\,\cos \,\theta \,2u\,\sin \,\theta }}{{g\,\cos \,\alpha }} - \frac{{g\,\sin \,\alpha }}{2}\frac{{4{u^2}{{\sin }^2}\theta }}{{{g^2}{{\cos }^2}\alpha }}\\
\,\,\,\,\, = \frac{{{u^2}{{\sin }^2}\theta }}{{g\,\cos \,\alpha }} - \frac{{{u^2}\sin \,\alpha }}{{g\,{{\cos }^2}\alpha }}\{ 1 - \cos 2\theta \} 
\end{array}\)

\(\begin{array}{l}
 = \frac{{4 \times \frac{1}{2}}}{{10 \times \frac{{\sqrt 3 }}{2}}} - \frac{{{u^2}\sin \,\alpha }}{{g\,{{\cos }^2}\,\alpha }}\left\{ {1 - \frac{{\sqrt 3 }}{2}} \right\}\\
 = \frac{4}{{10\sqrt 3 }} - \frac{8}{{30}}\left\{ {1 - \frac{{\sqrt 3 }}{2}} \right\}\\
 = \frac{4}{{5\sqrt 3 }} - \frac{8}{{30}} = \frac{{8\sqrt 3  - 8}}{{30}} = \frac{{8\left( {\sqrt 3  - 1} \right)}}{{30}} = 20\,cm
\end{array}\)