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(C) Given: Horizontal range $(R)$ $= 2 	imes$ Maximum height $(H)$.Formula for range: $R = \frac{v^{2} \sin 2	heta}{g}$.Formula for maximum height:

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

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(C) Given: Horizontal range $(R)$ $= 2 	imes$ Maximum height $(H)$.Formula for range: $R = \frac{v^{2} \sin 2	heta}{g}$.Formula for maximum height: $H = \frac{v^{2} \sin^{2} 	heta}{2g}$.According to the problem: $\frac{v^{2} \sin 2	heta}{g} = 2 	imes \frac{v^{2} \sin^{2} 	heta}{2g}$.Simplifying: $\sin 2	heta = \sin^{2} 	heta$.Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$: $2 \sin 	heta \cos 	heta = \sin^{2} 	heta$.Dividing by $\sin 	heta$ (assuming $	heta 
eq 0$): $2 \cos 	heta = \sin 	heta$,which means $	an 	heta = 2$.From $	an 	heta = 2$,we have $\sin 	heta = \frac{2}{\sqrt{5}}$ and $\cos 	heta = \frac{1}{\sqrt{5}}$.Substituting these into the range formula: $R = \frac{v^{2} (2 \sin 	heta \cos 	heta)}{g} = \frac{2v^{2}}{g} 	imes \frac{2}{\sqrt{5}} 	imes \frac{1}{\sqrt{5}} = \frac{4v^{2}}{5g}$.

Column $I$	Column $II$
$(A)$ Horizontal component of velocity	$(p)$ $5$ $SI$ unit
$(B)$ Vertical component of velocity	$(q)$ $10$ $SI$ unit
$(C)$ Horizontal displacement	$(r)$ $15$ $SI$ unit
$(D)$ Vertical displacement	$(s)$ $20$ $SI$ unit

$A$ particle is projected with a velocity $v$ such that its horizontal range is twice the greatest height attained. The horizontal range is

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