PrepRung
Sign inSign up free

JEE Main 2018 · Physics · Kinematics and Laws of Motion · hard

A projectile is launched from the foot of a smooth fixed incline of angle 30^° with speed 15 m s^-1, at 30^° above the incline surface (so the…

Single correct option

A projectile is launched from the foot of a smooth fixed incline of angle 30∘30^\circ with speed 15 m s−115\ \mathrm{m\,s^{-1}}, at 30∘30^\circ above the incline surface (so the velocity makes 60∘60^\circ with the horizontal and lies in the vertical plane containing the line of greatest slope). Taking g=10 m s−2g=10\ \mathrm{m\,s^{-2}}, the distance along the incline at which it lands again on the incline is
  1. A. 30 m30\ \mathrm{m}
  2. B. 19.5 m19.5\ \mathrm{m}
  3. C. 13.0 m13.0\ \mathrm{m}
  4. D. 15 m15\ \mathrm{m}
Show answer and solution

Answer: (D) 15 m15\ \mathrm{m}

Take axes along and perpendicular to the incline. Then u∥=ucos⁡30∘u_\parallel=u\cos30^\circ, u⊥=usin⁡30∘u_\perp=u\sin30^\circ, and the components of gravity are g∥=−gsin⁡30∘=−5g_\parallel=-g\sin30^\circ=-5, g⊥=−gcos⁡30∘=−53g_\perp=-g\cos30^\circ=-5\sqrt3. Time of flight T=2u⊥gcos⁡30∘=2(7.5)53=3 sT=\dfrac{2u_\perp}{g\cos30^\circ}=\dfrac{2(7.5)}{5\sqrt3}=\sqrt3\ \mathrm{s}. Distance along the incline =u∥T−12(5)T2=1532⋅3−52(3)=22.5−7.5=15 m=u_\parallel T-\tfrac12(5)T^2=\dfrac{15\sqrt3}{2}\cdot\sqrt3-\dfrac{5}{2}(3)=22.5-7.5=15\ \mathrm{m}. Using the horizontal-plane range (19.5 m19.5\ \mathrm{m}) ignores the incline.
Report a mistake

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake