A projectile is thrown at an angle theta such | vedclass

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(C) The projectile reaches its maximum height $H$ at a horizontal distance $x = \sqrt{3} H$ from the point of projection.For a projectile,the horizont

Vedclass · Accepted Answer

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

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(C) The projectile reaches its maximum height $H$ at a horizontal distance $x = \sqrt{3} H$ from the point of projection.For a projectile,the horizontal distance to the highest point is half the range,i.e.,$x = R/2$.We know the formulas for range $R$ and maximum height $H$:$R = \frac{v_0^2 \sin(2	heta)}{g} = \frac{2 v_0^2 \sin 	heta \cos 	heta}{g}$$H = \frac{v_0^2 \sin^2 	heta}{2g}$Given $x = R/2 = \sqrt{3} H$,we have:$\frac{R}{2H} = \sqrt{3}$Substituting the expressions for $R$ and $H$:$\frac{\frac{2 v_0^2 \sin 	heta \cos 	heta}{g}}{2 \left(\frac{v_0^2 \sin^2 	heta}{2g}ight)} = \sqrt{3}$$\frac{2 v_0^2 \sin 	heta \cos 	heta}{g} \cdot \frac{g}{v_0^2 \sin^2 	heta} = \sqrt{3}$$2 \cot 	heta = \sqrt{3}$$\cot 	heta = \frac{\sqrt{3}}{2}$$	an 	heta = \frac{2}{\sqrt{3}}$Therefore,$	heta = 	an ^{-1}\left(\frac{2}{\sqrt{3}}ight)$.

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