Q. 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) $v\cos\theta$
Explanation
(c) Average velocity $= \frac{displacement}{time}$
$V_{av} = \frac{\sqrt{H^{2} + \frac{R^{2}}{4}}}{T/2}$ ….(i)
[image:4_image147.emf]
Here $H =$maximum height $= \frac{v^{2}\sin^{2}\theta}{2g}$
$R =$range $= \frac{v^{2}{sin\ }{2\theta}}{g}$
and $T =$time of flight $= \frac{2v{sin\ }\theta}{g}$
Substituting in Eq. (i), we get
$v_{av} = \frac{v}{2}\sqrt{1 + 3\cos^{2}\theta}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.