If R and H are the horizontal range and maxim | vedclass

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Question

(b)

$H=\frac{u^2 \sin ^2 	heta}{2 g} \Rightarrow \sin 	heta=\sqrt{\frac{2 g H}{u^2}}$

$R=\frac{2 u^2 \sin 	heta \cos 	heta}{g}$

$R=\frac{

Vedclass · Accepted Answer

$\sqrt{2 g H+\frac{R^2 g}{8 H}}$

Vedclass · Answer

(b)

$H=\frac{u^2 \sin ^2 	heta}{2 g} \Rightarrow \sin 	heta=\sqrt{\frac{2 g H}{u^2}}$

$R=\frac{2 u^2 \sin 	heta \cos 	heta}{g}$

$R=\frac{2 u^2}{g} \sqrt{\frac{2 g H}{u^2}} 	imes \sqrt{1-\frac{2 g H}{u^2}}$

$R=\frac{2 u^2}{g} \sqrt{\frac{2 g H}{u^2}} 	imes \sqrt{\frac{u^2-2 g H}{u^2}}$

$\frac{g R}{2 \sqrt{2 g H}}=\sqrt{u^2-2 g H}$

Squaring both the sides,

$\frac{g R^2}{4 	imes 2 g H}=u^2-2 g H$

$\Rightarrow u^2=2 g H+\frac{9 R^2}{8 H}$

$u=\sqrt{2 g H+\frac{g R^2}{8 H}}$

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

If $R$ and $H$ are the horizontal range and maximum height attained by a projectile, than its speed of projection is ..........

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Match the columns

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

$A.$ $1$ $1.$ ${60^o}$

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

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

$D.$ $\frac {4}{\sqrt 3}$ $4.$ $tan^{-1}\,4\,=\,{76^o}$