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AP EAMCET201823 Apr 2018Evening ShiftPhysicsMotion in Two DimensionsActual

A particle is projected at an angle of 60^ with the horizontal from the ground with a velocity 10 3 ~ms ⁻¹ . The angle between velocity vector after 2 ~s and initial velocity vector is (g=10 ~ms ⁻² )

Options

  1. A0
  2. B30^
  3. C60^
  4. D90^

Correct answer

D. 90^

Step-by-step solution

Initial velocity, v _ i =2 i +4 j =5 3 i +15 j Final velocity vector (after 2 s ), v _ f =u i +(u -g t) j =5 3 i -5 j Now, v _ i v _ f =25 3-15 5=0 v _ i v _ f

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