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

• A

  $u \cos \theta$
• B

  $\frac{u}{2} \sqrt{1+\cos ^2 \theta}$
• C

  $\frac{u}{2} \sqrt{1+2 \cos ^2 \theta}$
• Correct answer
  D

  $\frac{u}{2} \sqrt{1+3 \cos ^2 \theta}$

Solution

Refer figure as shown, the average velocity
$=\frac{\text { displacement }}{\text { time taken }}$ $v_{\mathrm{av}}=\frac{\sqrt{H^2+R^2 / 4}}{T / 2}$...(i) Here, $H=\frac{u^2 \sin ^2 \theta}{2 g}$ $R=\frac{u^2 \sin 2 \theta}{g}$ Putting these values in Eq. (i), we get $v_{\mathrm{av}}=\frac{u}{2} \sqrt{1+3 \cos ^2 \theta}$

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