Question

A particle is projected with a velocity of $\sqrt{7}m{s}^{-1}$ at an 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. $2m/s$
• B. $\frac{5}{4}m/s$
• C. $\frac{7}{4}m/s$
• D. $\frac{3}{4}m/s$

Answer 1

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

Given $u=\sqrt{7}m{s}^{-1},\theta ={60}^{\circ }\text{Average velocity}=\frac{\text{Total displacement}}{\text{time}}$
From the figure, $\Delta r$ is the displacement of the particle.
$\Rightarrow v_{avg}=\frac{△r}{\frac{T}{2}}=\frac{\sqrt{H^2+\frac{R^2}{4}}}{\frac{T}{2}}$
But, $\frac{H}{R}=\frac{\tan \theta }{4}\Rightarrow H=\frac{R\tan \theta }{4}\Rightarrow v_{avg}=\frac{\sqrt{\frac{R^2{\tan }^2\theta +}{16}\frac{R^2}{4}}}{\frac{T}{2}}=\frac{\sqrt{\frac{R^2}{4}}}{\frac{T}{2}}\sqrt{1+\frac{{\tan }^2\theta }{4}}=\frac{2u^2\sin \theta \cos \theta }{2u\sin \theta }\sqrt{\frac{4+{\tan }^2\theta }{4}}=\frac{u}{2}\sqrt{4{\cos }^2\theta +{\sin }^2\theta }=\frac{u}{2}\sqrt{1+3{\cos }^2\theta }$
On puting the values of $u$ and $\theta$ in above equation, we get
$v_{avg}=\frac{\sqrt{7}}{2}\sqrt{1+3\times \frac{1}{4}}\Rightarrow v_{avg}=\frac{7}{4}m{s}^{-1}$