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

• A

  $\frac{v}{2} \sqrt{1+2 \cos ^2 \theta}$
• B

  $\frac{v}{2} \sqrt{1+\cos ^2 \theta}$
• Correct answer
  C

  $\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}$
• D

  $v \cos \theta$

Hence, the correct answer is option: 'C'.