Question

A particle is projected from the bottom of an inclined plane of angle ${30}^{\circ }$ with a velocity of $20\text{m/s}$ at an angle of ${45}^{\circ }$ with the horizontal. Find the horizontal distance travelled by the particle before it hits the plane.

• A. $x=\frac{80}{\sqrt{3}}[\sqrt{3}-1]\text{m}$
• B. $x=\frac{40}{\sqrt{3}}[\sqrt{3}-1]\text{m}$
• C. $x=\frac{40}{3}[\sqrt{3}-1]\text{m}$
• D. $x=\frac{80}{3}[\sqrt{3}-1]\text{m}$

Answer 1

The correct option is B $x=\frac{40}{\sqrt{3}}[\sqrt{3}-1]\text{m}$

Horizontal distance travelled by the particle is $x=R\cos \theta ,$ where $R$ is the range of a projectile on an inclined plane.
We know $R=\frac{u^2}{g{\cos }^2\theta }\left[\sin \right(2a-\theta )-\sin \theta ]\Rightarrow R=\frac{{20}^2}{10\times {\cos }^2{30}^{\circ }}\left[\sin \right(2\times {45}^{\circ }-{30}^{\circ })-\sin ({30}^{\circ }\left)\right]\Rightarrow R=\frac{400}{10\times \frac{3}{4}}[\frac{\sqrt{3}}{2}-\frac{1}{2}]\Rightarrow R=\frac{80}{3}[\sqrt{3}-1]$

Hence, $x=R\cos {30}^{\circ }$
$=\frac{40}{\sqrt{3}}[\sqrt{3}-1]\text{m}$