Question

Single Correct Answer Type
A particle is projected from the ground with an initial speed of 'v' at angle $\theta$ with horizontal. the average velocity of the particle between its point of projection and height 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)

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

From figure,
average velocity,
$u_{av}=\frac{\sqrt{H^2+R^2/4}}{T/2}$ ...... (i)

Here, $H=\frac{u^2{\sin }^2\theta }{2g}$
$R=\frac{u^2\sin \theta }{g}$ and $T=\frac{2u\sin \theta }{g}$

Putting these value in (i), we get
$u_{av}=\frac{v}{2}\sqrt{1+3{\cos }^2\theta }$