A projectile can have the same range for two | vedclass

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Question

$\begin{array}{l}
{h_1} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}\
{h_2} = \frac{{{u^2}{{\sin }^2}\left( {90 - 	heta } ight)}}{{2g}},R = \frac{{{

Vedclass · Accepted Answer

$R = 4\sqrt {{h_1}{h_2}} $

Vedclass · Answer

$\begin{array}{l}
{h_1} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}\
{h_2} = \frac{{{u^2}{{\sin }^2}\left( {90 - 	heta } ight)}}{{2g}},R = \frac{{{u^2}\sin 2	heta }}{g}\
Range\,R\,is\,same\,for\,angle\,	heta \,and\,\left( {{{90}^ \circ } - 	heta } ight)\
	herefore \,{h_1}{h_2} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}} 	imes \frac{{{u^2}{{\sin }^2}\left( {90 - 	heta } ight)}}{{2g}}\
 = \frac{{{u^4}\left( {{{\sin }^2}	heta } ight) 	imes {{\sin }^2}\left( {90 - 	heta } ight)}}{{4{g^2}}}\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\sin \left( {90 - 	heta } ight) = 	heta \cos 	heta } ight]
\end{array}$
$\begin{array}{l}
 = \frac{{{u^4}{{\left( {\sin 	heta \cos 	heta } ight)}^2} 	imes {{\cos }^2}	heta }}{{4{g^2}}}\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\sin 2	heta  = 2\sin 	heta \cos 	heta } ight]\
 = \frac{{{u^4}{{\left( {\sin 	heta \cos 	heta } ight)}^2}}}{{4{g^2}}} = \frac{{{u^4}{{\left( {\sin 2	heta } ight)}^2}}}{{16g}}\
 = \frac{{{{\left( {{u^2}\sin 2	heta } ight)}^2}}}{{16{g^2}}} = \frac{{{R^2}}}{{16}}\
or,\,{R^2} = 16{h_1}{h_2}\,or\,R = 4\sqrt {{h_1}{h_2}} 
\end{array}$

A projectile can have the same range for two angles of projection. If $h_1$ and $h_2$ are maximum heights when the range in the two cases is $R$, then the relation between $R$, $h_1$ and $h_2$ is

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