Question

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

Answer 1

Answer: (C)

We know, average velocity
$=\frac{\text { displacement }}{\text { time }}$
$v_{ av }=\frac{\sqrt{\mu^{2}+R^{2} / 4}}{T / 2}$ ...(i)

where, $H=$ maximum height
$=\frac{v^{2} \sin ^{2} \theta}{2 g}$ ... (ii)
Range $R=\frac{v^{2} \sin 2 \theta}{g}$ ...(iii)
Time of flight $T=\frac{2 v \sin \theta}{g}$
Putting the values of Eqs. (ii), (iii) and (iv) in Eq. (i) we have
$v_{ av }=\frac{v}{2} \sqrt{1+3 \cos ^{2} \theta}$