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NEET2013PhysicsMotion in Two DimensionsActual

A particle is projected from the ground with an initial speed of ' v ' at angle with horizontal. The average velocity of the particle between its point of projection and height point of trajectory is

Options

  1. Av 2 1+2 ^2
  2. Bv 2 1+ ^2
  3. Cv 2 1+3 ^2
  4. Dv

Correct answer

C. v 2 1+3 ^2

Step-by-step solution

From figure,average velocity, v_ av = H^2+R^2 / 4 T / 2 ( i ) Here, H= u^2 ^2 2 g R= u^2 2 g and T= 2 u g Putting these value in (i), we get v_ av = v 2 1+3 ^2

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