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AP EAMCET2013PhysicsMotion in Two Dimensions

A particle is projected from the ground with an initial speed of v at an angle of projection . The average velocity of the particle between its time of projection and time it reaches highest point of trajectory is

Options

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

Correct answer

C. v 2 1+3 ^2

Step-by-step solution

aligned & We know, average velocity = displacement time & v_ av = ^2+R^2 / 4 T / 2 aligned where, H= maximum height = v^2 ^2 2 g Range R= v^2 2 g Time of flight T= 2 v g Putting the values of Eqs. (ii), (iii) and (iv) in Eq. (i) we have v_ av = v 2 1+3 ^2

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