A particle is projected with a velocity v suc | vedclass

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Question

(a) $R = 2H$ given

We know $R = 4H\cot 	heta $==> $\cot 	heta = \frac{1}{2}$

From triangle we can say that $\sin 	heta = \frac{2}{{\sqrt 5

Vedclass · Accepted Answer

$\frac{{4{v^2}}}{{5g}}$

Vedclass · Answer

(a) $R = 2H$ given

We know $R = 4H\cot 	heta $==> $\cot 	heta = \frac{1}{2}$

From triangle we can say that $\sin 	heta = \frac{2}{{\sqrt 5 }}$,$\cos 	heta = \frac{1}{{\sqrt 5 }}$

$	herefore$ Range of projectile $R = \frac{{2{v^2}\sin 	heta \cos 	heta }}{g}$

$ = \frac{{2{v^2}}}{g} 	imes \frac{2}{{\sqrt 5 }} 	imes \frac{1}{{\sqrt 5 }}$ = $\frac{{4{v^2}}}{{5g}}$.

A particle is projected with a velocity $v$ such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where $g$ is acceleration due to gravity)

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