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Question

Average velocity $=\frac{	ext { Displacement }}{	ext { Time }}$

$v_{w_{w}}=\frac{\sqrt{H^{2}+\frac{R^{2}}{4}}}{T / 2}$

Here, $\quad \mathrm{H}

Vedclass · Accepted Answer

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

Vedclass · Answer

Average velocity $=\frac{	ext { Displacement }}{	ext { Time }}$

$v_{w_{w}}=\frac{\sqrt{H^{2}+\frac{R^{2}}{4}}}{T / 2}$

Here, $\quad \mathrm{H}=$ maximum height $=\frac{v^{2} \sin ^{2} 	heta}{2 g}$

$\mathrm{R}=\mathrm{range}=\frac{\mathrm{v}^{2} \sin 2 	heta}{\mathrm{g}}$

and $\quad \mathrm{T}=$ time of flight $=\frac{2v \sin 	heta}{\mathrm{g}}$

$	herefore \quad \mathrm{U}_{\mathrm{av} .}=\frac{\mathrm{v}}{2} \sqrt{1+3 \cos ^{2} 	heta}$

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

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