The horizontal range and the maximum height o | vedclass

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Question

$\begin{array}{l}
Horizontal\,range,\,R = \frac{{{u^2}\sin 2	heta }}{g}\
Where\,u\,is\,the\,velocity\,of\,projection\,and\,	heta \
is\,the\,ang

Vedclass · Accepted Answer

$	heta = {	an ^{ - 1}}4$

Vedclass · Answer

$\begin{array}{l}
Horizontal\,range,\,R = \frac{{{u^2}\sin 2	heta }}{g}\
Where\,u\,is\,the\,velocity\,of\,projection\,and\,	heta \
is\,the\,angle\,of\,projection\
Maximum\,height,\,H = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}\
According\,to\,question\,R = H
\end{array}$

$\begin{array}{l}
	herefore \,\,\frac{{{u^2}\sin 2	heta }}{g} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}\
\,\,\,\,\frac{{2{u^2}\sin 	heta \cos 	heta }}{g} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}\
\,\,	an 	heta  = 4\,\,or\,	heta  = {	an ^{ - 1}}\left( 4 ight)
\end{array}$

The horizontal range and the maximum height of a projectile are equal . The angle of projection of the projectile is

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