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JEE Main 2024 · Physics · Kinematics and Laws of Motion · medium

A particle is projected from a point on a smooth-surfaced incline of inclination 30^° with speed 20 m s^-1, in a direction perpendicular to the…

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A particle is projected from a point on a smooth-surfaced incline of inclination 30∘30^\circ with speed 20 m s−120\ \mathrm{m\,s^{-1}}, in a direction perpendicular to the incline. Taking g=10 m s−2g=10\ \mathrm{m\,s^{-2}}, the distance along the incline from the point of projection to the point where it lands is nearly
  1. A. 46.2 m46.2\ \mathrm{m}
  2. B. 40.0 m40.0\ \mathrm{m}
  3. C. 80.0 m80.0\ \mathrm{m}
  4. D. 53.3 m53.3\ \mathrm{m}
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Answer: (D) 53.3 m53.3\ \mathrm{m}

Resolve axes along and perpendicular to the incline. Perpendicular: initial speed 2020, deceleration gcos⁡30∘=53g\cos30^\circ=5\sqrt3, so the time of flight is T=2⋅2053=83 sT=\dfrac{2\cdot20}{5\sqrt3}=\dfrac{8}{\sqrt3}\ \mathrm{s}. Along the incline: initial speed 00, acceleration gsin⁡30∘=5g\sin30^\circ=5 (down the slope). Distance =12(5)T2=12(5)643=1603≈53.3 m=\tfrac12(5)T^2=\tfrac12(5)\dfrac{64}{3}=\dfrac{160}{3}\approx53.3\ \mathrm{m}.
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