Question:
A ball is projected from the point O with velocity 20 m/s at an angle of 60° with horizontal. At highest point of its trajectory, it strikes a smooth plane of inclination 30° at point A. The collision is perfectly inelastic. The maximum height from the ground attained by the ball is
A ball is projected from the point O with velocity 20 m/s at an angle of 60° with horizontal. At highest point of its trajectory, it strikes a smooth plane of inclination 30° at point A. The collision is perfectly inelastic. The maximum height from the ground attained by the ball is
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In perfectly inelastic collision with a smooth plane, the component of velocity parallel to the plane is conserved.
Updated On: Apr 20, 2026
- 18.75 m
- 15 m
- 22.5 m
- 20.25 m
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
At highest point, velocity is horizontal: \(v = u\cos\theta = 20\cos60^\circ = 10 \text{ m/s}\). In perfectly inelastic collision with incline: velocity component along the plane remains, perpendicular component becomes zero.
Step 2: Detailed Explanation:
Height of highest point \(A\): \(H_A = \frac{u^2\sin^2\theta}{2g} = \frac{400 \times 0.75}{20} = 15 \text{ m}\). Velocity along incline after collision: \(v = 10\cos30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ m/s}\). Additional height along incline: \(h = \frac{v^2}{2g\sin30^\circ} = \frac{75}{2 \times 10 \times 0.5} = \frac{75}{10} = 7.5 \text{ m}\). Vertical height gained: \(7.5\sin30^\circ = 3.75 \text{ m}\). Total maximum height: \(15 + 3.75 = 18.75 \text{ m}\).
Step 3: Final Answer:
\[ \boxed{18.75 \text{ m}} \]
At highest point, velocity is horizontal: \(v = u\cos\theta = 20\cos60^\circ = 10 \text{ m/s}\). In perfectly inelastic collision with incline: velocity component along the plane remains, perpendicular component becomes zero.
Step 2: Detailed Explanation:
Height of highest point \(A\): \(H_A = \frac{u^2\sin^2\theta}{2g} = \frac{400 \times 0.75}{20} = 15 \text{ m}\). Velocity along incline after collision: \(v = 10\cos30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ m/s}\). Additional height along incline: \(h = \frac{v^2}{2g\sin30^\circ} = \frac{75}{2 \times 10 \times 0.5} = \frac{75}{10} = 7.5 \text{ m}\). Vertical height gained: \(7.5\sin30^\circ = 3.75 \text{ m}\). Total maximum height: \(15 + 3.75 = 18.75 \text{ m}\).
Step 3: Final Answer:
\[ \boxed{18.75 \text{ m}} \]
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