एक प्रक्षेप्य को क्षैतिज से 45^{circ} के कोण | vedclass

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Question

$\begin{array}{l}
Let\,\phi \,be\,elevation\,angle\,of\,the\,projectile\
at\,its\,highest\,{m{point}}\,{m{as}}\,{m{seen}}\,{m{from}}\,{m{

Vedclass · Accepted Answer

$ta{n^{ - 1}}\left( {\frac{1}{2}} \right)$

Vedclass · Answer

$\begin{array}{l}
Let\,\phi \,be\,elevation\,angle\,of\,the\,projectile\
at\,its\,highest\,{m{point}}\,{m{as}}\,{m{seen}}\,{m{from}}\,{m{the}}\
{m{ point}}\,of\,projection\,O\,and\,	heta \,be\,angle\,of\
projection\,with\,the\,horizontal.\
From\,figure,\,	an \phi  = \frac{H}{{R/2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)\
In\,case\,of\,projectile\,motion\
Maximum\,height,\,H = \frac{{{h^2}{{\sin }^2}2	heta }}{{2g}}\
Horizontal\,range,R = \frac{{{h^2}\sin 2	heta }}{g}\
Substituting\,theses\,values\,of\,H\,and\,R\
in\,\left( i ight),\,we\,get
\end{array}$

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,	an \phi  = \frac{{\frac{{{u^2}{{\sin }^2}	heta }}{{2g}}}}{{\frac{{{u^2}\sin 	heta }}{{2g}}}}\
\,\,\,\,\,\,\,\,\,\,	an \phi  = \frac{{{{\sin }^2}	heta }}{{\sin 2	heta }} = \frac{{{{\sin }^2}	heta }}{{2\sin 	heta \cos 	heta }} = \frac{1}{2}	an 	heta \
\,\,\,\,	an \phi  = \frac{1}{2}	an {45^ \circ } = \frac{1}{2}\
Here,	heta  = {45^ \circ }\
	herefore \,\,\,\,\,	an \phi  = \frac{1}{2}	an {45^ \circ } = \frac{1}{2}\,\,\,\,\left( {\,	an \,{{45}^ \circ } = 1} ight)\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\phi  = {	an ^{ - 1}}\left( {\frac{1}{2}} ight)
\end{array}$

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