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

The correct option is C $\frac{v}{2}\sqrt{1+3{\cos }^2\theta }$



Average velocity = $\frac{Displacement}{Time}$
$=\frac{\sqrt{H^2+\frac{R^2}{4}}}{\frac{T}{2}}$

we know that, $H=\frac{v^2{\sin }^2\theta }{2g}$, $R=\frac{v^2\sin 2\theta }{g}$, $T=\frac{2v\sin \theta }{g}$

$v_{av}=\left(\frac{g}{v\sin \theta }\right)\left(\sqrt{(\frac{v^2{\sin }^2\theta }{2g}{)}^2+\frac{(v^2\sin 2\theta {)}^2}{4g^2}}\right)$
$=\sqrt{\frac{v^4{\sin }^4\theta +v^4{\sin }^{2}2\theta }{4g^2}\times \frac{g^2}{v^2{\sin }^2\theta }}$

On simplifying further,
$v_{av}=\frac{v}{2}\sqrt{1+3{\cos }^2\theta }$