For angles of projection of a projectile at a | vedclass

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Question

$\begin{array}{l}
Horizontal\,range{\kern 1pt} R = \frac{{{u^2}\sin 2	heta }}{g}\
For\,angle\,of\,projection\,\left( {{{45}^ \circ } - 	heta }

Vedclass · Accepted Answer

$1:1$

Vedclass · Answer

$\begin{array}{l}
Horizontal\,range{\kern 1pt} R = \frac{{{u^2}\sin 2	heta }}{g}\
For\,angle\,of\,projection\,\left( {{{45}^ \circ } - 	heta } ight),\,the\
horizontal\,range\,is\,\
	herefore \,{R_1} = \frac{{{u^2}\sin \left[ {2\left( {{{45}^ \circ } - 	heta } ight)} ight]}}{g}\
\,\,\,\,\,\,\,\,\,\,\, = \frac{{{u^2}\,\sin \left( {{{90}^ \circ } - 2	heta } ight)}}{g}\
\,\,\,\,\,\,\,\,\,\,\, = \frac{{{u^2}\cos \,2	heta }}{g}
\end{array}$

$\begin{array}{l}
For\,angle\,of\,projection\,\left( {{{45}^ \circ } + 	heta } ight),\,the\
horizontal\,range\,is\
{R_2} = \frac{{{u^2}\sin \left[ {2\left( {{{45}^ \circ } + 	heta } ight)} ight]}}{g}\
\,\,\,\,\,\, = \frac{{{u^2}\,\sin \left( {{{90}^ \circ } + 2	heta } ight)}}{g} = \frac{{{u^2}\cos \,2	heta }}{g}\
	herefore \,\,\,\,\frac{{{R_1}}}{{{R_2}}} = \frac{{{u^2}\cos 2	heta /g}}{{{u^2}\cos 2	heta /g}} = \frac{1}{1}.\
	herefore \,The\,range\,is\,the\,same.
\end{array}$

For angles of projection of a projectile at angle $(45^o +\theta)$ and $(45^o -\theta ) $ , the horizontal range described by the projectile are in the ratio of

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