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(C) The average velocity is defined as the total displacement divided by the total time taken.Let the particle be projected from the origin $(0,0)$. A

Vedclass · Accepted Answer

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

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(C) The average velocity is defined as the total displacement divided by the total time taken.Let the particle be projected from the origin $(0,0)$. At the highest point, the coordinates are $(R/2, H)$, where $R$ is the horizontal range and $H$ is the maximum height.Displacement vector $\vec{s} = \frac{R}{2} \hat{i} + H \hat{j}$.The magnitude of displacement is $|\vec{s}| = \sqrt{(R/2)^2 + H^2}$.The time taken to reach the highest point is $t = \frac{T}{2} = \frac{v \sin 	heta}{g}$.We know $R = \frac{v^2 \sin 2	heta}{g} = \frac{2v^2 \sin 	heta \cos 	heta}{g}$ and $H = \frac{v^2 \sin^2 	heta}{2g}$.Thus, $R/2 = \frac{v^2 \sin 	heta \cos 	heta}{g}$.$|\vec{s}| = \sqrt{\left(\frac{v^2 \sin 	heta \cos 	heta}{g}ight)^2 + \left(\frac{v^2 \sin^2 	heta}{2g}ight)^2} = \frac{v^2 \sin 	heta}{g} \sqrt{\cos^2 	heta + \frac{\sin^2 	heta}{4}} = \frac{v^2 \sin 	heta}{2g} \sqrt{4 \cos^2 	heta + \sin^2 	heta} = \frac{v^2 \sin 	heta}{2g} \sqrt{3 \cos^2 	heta + 1}$.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 of $v$ at an angle of projection $\theta$. The average velocity of the particle between its time of projection and the time it reaches the highest point of its trajectory is

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