Question

A particle is projected with a velocity of $\sqrt{7}{\text{ms}}^{-1}$ at angle ${60}^{\circ }$ with horizontal. Find the average velocity of the projectile between the instants of point of projection and reaching the highest point.

• A. $2\text{m/s}$
• B. $\frac{5}{4}\text{m/s}$
• C. $\frac{7}{4}\text{m/s}$
• D. $\frac{3}{4}\text{m/s}$

Answer 1

The correct option is C $\frac{7}{4}\text{m/s}$

Given $u=\sqrt{7}{\text{ms}}^{-1},\theta ={60}^{\circ }$
$\text{Average velocity}=\frac{\text{Total displacement}}{\text{Time}}$

From the figure, $\Delta r$ is the displacement of the particle.
$\Rightarrow v_{avg}=\frac{\Delta r}{T}=\frac{\sqrt{H^2+\frac{R^2}{4}}}{\frac{T}{2}}$
But $\frac{H}{R}=\frac{\tan \theta }{4}$ or $H=\frac{R\tan \theta }{4}$
$\Rightarrow v_{avg}=\frac{\sqrt{\frac{R^2{\tan }^2\theta }{16}+\frac{R^2}{4}}}{\frac{T}{2}}$
$=\frac{\sqrt{\frac{R^2}{4}}}{\frac{T}{2}}\sqrt{1+\frac{{\tan }^2\theta }{4}}$
$=\frac{2u^2\sin \theta \cos \theta }{2u\sin \theta }\sqrt{\frac{4+{\tan }^2\theta }{4}}$
$=\frac{u}{2}\sqrt{4{\cos }^2\theta +{\sin }^2\theta }$
$=\frac{u}{2}\sqrt{1+3{\cos }^2\theta }$
On putting the values of $u$ and $\theta$ in above equation, we get
$v_{avg}=\frac{\sqrt{7}}{2}\sqrt{1+3\times \frac{1}{4}}$
$\Rightarrow v_{avg}=\frac{7}{4}{\text{ms}}^{-1}$
Hence, the correct answer is option (c)