Match the columns: Column-I (R/H_{max}) | vedclass

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(B) The range $R$ and maximum height $H$ of a projectile are given by:$R = \frac{u^2 \sin(2	heta)}{g} = \frac{2u^2 \sin	heta \cos	heta}{g}$$H = \fr

Vedclass · Accepted Answer

$A-4, B-3, C-2, D-1$

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(B) The range $R$ and maximum height $H$ of a projectile are given by:$R = \frac{u^2 \sin(2	heta)}{g} = \frac{2u^2 \sin	heta \cos	heta}{g}$$H = \frac{u^2 \sin^2	heta}{2g}$Taking the ratio $R/H$:$\frac{R}{H} = \frac{2u^2 \sin	heta \cos	heta}{g} 	imes \frac{2g}{u^2 \sin^2	heta} = 4 \cot	heta$Therefore,$\frac{R}{H} = 4 \cot	heta$ or $	an	heta = \frac{4H}{R}$.For $A. R/H = 1$: $1 = 4 \cot	heta \implies 	an	heta = 4 \implies 	heta = 	an^{-1}(4) = 76^o$ (Matches $4$).For $B. R/H = 4$: $4 = 4 \cot	heta \implies 	an	heta = 1 \implies 	heta = 45^o$ (Matches $3$).For $C. R/H = 4\sqrt{3}$: $4\sqrt{3} = 4 \cot	heta \implies 	an	heta = 1/\sqrt{3} \implies 	heta = 30^o$ (Matches $2$).For $D. R/H = 4/\sqrt{3}$: $4/\sqrt{3} = 4 \cot	heta \implies 	an	heta = \sqrt{3} \implies 	heta = 60^o$ (Matches $1$).Thus,the correct matching is $A-4, B-3, C-2, D-1$.

Column-$I$ $(R/H_{max})$	Column-$II$ (Angle of projection $\theta$)
$A. 1$	$1. 60^o$
$B. 4$	$2. 30^o$
$C. 4\sqrt{3}$	$3. 45^o$
$D. 4/\sqrt{3}$	$4. \tan^{-1}(4) = 76^o$

Match the columns:

Column-$I$ $(R/H_{max})$ Column-$II$ (Angle of projection $\theta$)

$A. 1$ $1. 60^o$

$B. 4$ $2. 30^o$

$C. 4\sqrt{3}$ $3. 45^o$

$D. 4/\sqrt{3}$ $4. \tan^{-1}(4) = 76^o$

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Match the columns: Column-$I$ $(R/H_{max})$ Column-$II$ (Angle of projection $\theta$) $A. 1$ $1. 60^o$ $B. 4$ $2. 30^o$ $C. 4\sqrt{3}$ $3. 45^o$ $D. 4/\sqrt{3}$ $4. \tan^{-1}(4) = 76^o$