Projectile On An Inclined Plane

Projectile motion onto an incline is best handled by writing the horizontal and vertical displacements separately and imposing the geometric constraint that the landing point must lie on the sloping surface. Taking the launch point as origin with $x$ measured horizontally and $y$ measured vertically downward, the horizontal displacement is $x=ut=20t$ and the vertical drop under gravity is $y=\frac{1}{2}g{t}^2=5t^2$. Because the incline descends at 30 degrees below the horizontal, any point on its surface satisfies $\tan 30^{\circ }=\frac{y}{x}$, equating the ratio of vertical drop to horizontal run with the tangent of the slope angle. Substituting gives $\frac{5t^2}{20t}=\frac{1}{\sqrt{3}}$, which simplifies to $\frac{t}{4}=\frac{1}{\sqrt{3}}$, so $t=\frac{4}{\sqrt{3}}\text{ s}\approx 2.31\text{ s}$. The option $\frac{2}{\sqrt{3}}$ s halves the correct root by misplacing the factor of four from the displacement ratio. The option $\sqrt{3}$ s ignores the launch speed entirely. The option 2 s assumes a 45-degree slope rather than 30 degrees. This is the standard JEE Advanced slope-constraint method for projectiles. As a check, a gentler 30-degree incline lets the particle travel farther before meeting the surface, so a flight time above two seconds is physically reasonable.