A projectile is thrown at an angle theta such | vedclass

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Question

(c)

$\frac{R / 2}{H}=\frac{\sqrt{3} H}{H}=\sqrt{3}$

or $\frac{\left(v_0^2 \sin 	heta \cos 	hetaight) / g}{\left(v_0^2 \sin ^2 	hetaight)

Vedclass · Accepted Answer

$	an ^{-1}\left(\frac{2}{\sqrt{3}}ight)$

Vedclass · Answer

(c)

$\frac{R / 2}{H}=\frac{\sqrt{3} H}{H}=\sqrt{3}$

or $\frac{\left(v_0^2 \sin 	heta \cos 	hetaight) / g}{\left(v_0^2 \sin ^2 	hetaight) / 2 g}=\sqrt{3}$

$2 \cot 	heta=\sqrt{3}$

or $	an 	heta=\frac{2}{\sqrt{3}}$

or $	heta=	an ^{-1}\left(\frac{2}{\sqrt{3}}ight)$

A projectile is thrown at an angle $\theta$ such that it is just able to cross a vertical wall at its highest point as shown in the figure.The angle $\theta$ at which the projectile is thrown is given by

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