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

• A

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

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

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

  $v \cos \theta$

Solution

$\begin{aligned} & \text { We know, average velocity }=\frac{\text { displacement }}{\text { time }} \\ & \qquad v_{\text {av }}=\frac{\sqrt{\mu^2+R^2 / 4}}{T / 2}\end{aligned}$
where, $H=$ maximum height $=\frac{v^2 \sin ^2 \theta}{2 g}$ Range $R=\frac{v^2 \sin 2 \theta}{g}$ Time of flight $T=\frac{2 v \sin \theta}{g}$ Putting the values of Eqs. (ii), (iii) and (iv) in Eq. (i) we have $$ v_{\mathrm{av}}=\frac{v}{2} \sqrt{1+3 \cos ^2 \theta} $$

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