A plane is inclined at an angle $\alpha ={30}^{\circ }$ with respect to the horizontal. A particle is projected with a speed $u=2\text{m/s}$, from the base of the plane, as shown in figure. The distance from the base, at which the particle hits the plane, is close to:
(Take $g=10{\text{m/s}}^2$)

The correct option is A $20\text{cm}$
On an inclined plane, time of flight $\left(T\right)$ is given by,

$T=\frac{2u\sin \theta }{g\cos \alpha }$

Substituting the values, we get

$T=\frac{\left(2\right)\left(2\sin {15}^{\circ }\right)}{g\cos {30}^{\circ }}=\frac{4\sin {15}^{\circ }}{10\cos {30}^{\circ }}$

$\text{Distance, S}=\left(2\cos {15}^{\circ }\right)T-\frac{1}{2}g\sin {30}^{\circ }(T{)}^2$


$=\left(2\cos {15}^{\circ }\right)\frac{4}{10}\frac{\sin {15}^{\circ }}{10\cos {30}^{\circ }}-(\frac{1}{2}\times 10\sin {30}^{\circ })\frac{16{\sin }^2{15}^{\circ }}{100{\cos }^2{30}^{\circ }}$

$[\because \sin {15}^0=\frac{\sqrt{3}-1}{2\sqrt{2}}]$

$=\frac{16\sqrt{3}-16}{60}\simeq 0.1952\text{m}\simeq 20\text{cm}$

Hence, $\left(A\right)$ is the correct answer.