A particle is projected from the bottom of an inclined plane of angle ${30}^{\circ }$ with a velocity of $30\text{m/s}$ at an angle of ${60}^{\circ }$ with the horizontal. Find the distance travelled by the particle along the plane before it hits the plane.

The correct option is B $60\text{m}$
Distance travelled by the particle along the plane before hitting it is the range of the projectile on an inclined plane.
$\Rightarrow R=\frac{u^2}{g{\cos }^2\theta }\left[\sin \right(2\alpha -\theta )-\sin \theta ]$ (from the bottom of inclined plane)
$\Rightarrow R=\frac{{30}^2}{10\times {\cos }^2{30}^{\circ }}\left[\sin \right(2\times {60}^{\circ }-{30}^{\circ })-\sin {30}^{\circ }]\Rightarrow R=\frac{900}{10\times \frac{3}{4}}[1-\frac{1}{2}]\therefore R=60\text{m}$