Projectile On An Inclined Plane
A particle is projected with speed 20 m/s horizontally from the top of an incline that makes 30 degrees with the horizontal, directed down the slope. Taking g = 10 m/s^2, after what time does the particle strike the inclined surface?
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About This Question
- Subject
- physics
- Chapter
- kinematics
- Topic
- projectile on an inclined plane
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
4/√3s
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=21gt2=5t2. Because the incline descends at 30 degrees below the horizontal, any point on its surface satisfies tan30∘=xy, equating the ratio of vertical drop to horizontal run with the tangent of the slope angle. Substituting gives 20t5t2=31, which simplifies to 4t=31, so t=34 s≈2.31 s. The option 32 s halves the correct root by misplacing the factor of four from the displacement ratio. The option 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.
This hard difficulty physics question is from the chapter kinematics, covering the topic of projectile on an inclined plane. It appeared in the 2025 exam.
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