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(A) The time of flight for a projectile is given by $T = \frac{2v \sin 	heta}{g}$.For two complementary angles $	heta$ and $(90^\circ - 	heta)$,the

Vedclass · Accepted Answer

$250$

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(A) The time of flight for a projectile is given by $T = \frac{2v \sin 	heta}{g}$.For two complementary angles $	heta$ and $(90^\circ - 	heta)$,the horizontal range $R$ is the same.The times of flight are $T_1 = \frac{2v \sin 	heta}{g} = 5 	ext{ s}$ and $T_2 = \frac{2v \sin(90^\circ - 	heta)}{g} = \frac{2v \cos 	heta}{g} = 10 	ext{ s}$.The horizontal range is $R = \frac{v^2 \sin(2	heta)}{g} = \frac{2v^2 \sin 	heta \cos 	heta}{g}$.Multiplying $T_1$ and $T_2$,we get $T_1 T_2 = \frac{4v^2 \sin 	heta \cos 	heta}{g^2} = \frac{2}{g} \left( \frac{2v^2 \sin 	heta \cos 	heta}{g} ight) = \frac{2R}{g}$.Therefore,$R = \frac{1}{2} g T_1 T_2 = \frac{1}{2} 	imes 10 	imes 5 	imes 10 = 250 	ext{ m}$.

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

Two identical bodies,projected with the same speed at two different angles cover the same horizontal range $R$. If the time of flight of these bodies are $5 \text{ s}$ and $10 \text{ s}$,respectively,then the value of $R$ is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$)

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Match the columns:

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

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

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

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

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

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