Question

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

Answer 1

Answer: (C)

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

Answer 2

From figure, average velocity, $v_{ av }=\frac{\sqrt{H^{2}+R^{2} / 4}}{T / 2}$...(i) Here, $H=\frac{u^{2} \sin ^{2} \theta}{2 g}$ $R=\frac{u^{2} \sin 2 \theta}{g}$ and $T=\frac{2 u \sin \theta}{g}$ Putting these value in (i), we get $v_{ av }=\frac{v}{2} \sqrt{1+3 \cos ^{2} \theta}$