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(C) Average velocity is defined as the ratio of displacement to time.Let the point of projection be $(0, 0)$. The highest point of the trajectory is a

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Average velocity is defined as the ratio of displacement to time.Let the point of projection be $(0, 0)$. The highest point of the trajectory is at coordinates $(\frac{R}{2}, H)$,where $R$ is the horizontal range and $H$ is the maximum height.The displacement vector $\vec{s}$ from the origin to the highest point is $\vec{s} = \frac{R}{2}\hat{i} + H\hat{j}$.The magnitude of displacement is $|\vec{s}| = \sqrt{(\frac{R}{2})^2 + H^2}$.The time taken to reach the highest point is $t = \frac{T}{2} = \frac{v\sin	heta}{g}$.We know $H = \frac{v^2\sin^2	heta}{2g}$ and $R = \frac{v^2\sin2	heta}{g} = \frac{2v^2\sin	heta\cos	heta}{g}$.Thus,$\frac{R}{2} = \frac{v^2\sin	heta\cos	heta}{g}$.$|\vec{s}| = \sqrt{(\frac{v^2\sin	heta\cos	heta}{g})^2 + (\frac{v^2\sin^2	heta}{2g})^2} = \frac{v^2\sin	heta}{g} \sqrt{\cos^2	heta + \frac{\sin^2	heta}{4}} = \frac{v^2\sin	heta}{2g} \sqrt{4\cos^2	heta + \sin^2	heta}$.Using $\sin^2	heta = 1 - \cos^2	heta$,we get $|\vec{s}| = \frac{v^2\sin	heta}{2g} \sqrt{3\cos^2	heta + 1}$.Average velocity $v_{av} = \frac{|\vec{s}|}{t} = \frac{\frac{v^2\sin	heta}{2g} \sqrt{3\cos^2	heta + 1}}{\frac{v\sin	heta}{g}} = \frac{v}{2} \sqrt{1 + 3\cos^2	heta}$.

$A$ particle is projected from the ground with an initial speed $v$ at an angle $\theta$ with the horizontal. The average velocity of the particle between its point of projection and the highest point of its trajectory is

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