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NTA Abhyas JEE Main2020PhysicsMotion in Two DimensionsPractice

A ball is projected from point A with a velocity, 10 m s - 1 perpendicular to the inclined plane as shown in the figure. Range of the ball on the inclined plane is :

Options

  1. A4 0 3 m
  2. B2 0 13 m
  3. C13 20 m
  4. D13 40 m

Correct answer

A. 4 0 3 m

Step-by-step solution

T = 2u sin 90 o g cos 30 o = 2 u g cos 3 0 o R = 1 2 × g sin 30 o × 4 u 2 g 2 cos 2 30 o R = 2 u 2 × 4 2 × 1 0 × 3 = 2 × 1 0 × 1 0 × 4 2 × 1 0 × 3 = 4 0 3 m

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