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3-2.Motion in Plane
normal
The horizontal range and the maximum height of a projectile are equal . The angle of projection of the projectile is
A${45^o}$
B$\theta = {\tan ^{ - 1}}(0.25)$
C$\theta = {\tan ^{ - 1}}4$
D${60^o}$
Solution
$\begin{array}{l}
Horizontal\,range,\,R = \frac{{{u^2}\sin 2\theta }}{g}\\
Where\,u\,is\,the\,velocity\,of\,projection\,and\,\theta \\
is\,the\,angle\,of\,projection\\
Maximum\,height,\,H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\\
According\,to\,question\,R = H
\end{array}$
$\begin{array}{l}
\therefore \,\,\frac{{{u^2}\sin 2\theta }}{g} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\\
\,\,\,\,\frac{{2{u^2}\sin \theta \cos \theta }}{g} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\\
\,\,\tan \theta = 4\,\,or\,\theta = {\tan ^{ – 1}}\left( 4 \right)
\end{array}$
Horizontal\,range,\,R = \frac{{{u^2}\sin 2\theta }}{g}\\
Where\,u\,is\,the\,velocity\,of\,projection\,and\,\theta \\
is\,the\,angle\,of\,projection\\
Maximum\,height,\,H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\\
According\,to\,question\,R = H
\end{array}$
$\begin{array}{l}
\therefore \,\,\frac{{{u^2}\sin 2\theta }}{g} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\\
\,\,\,\,\frac{{2{u^2}\sin \theta \cos \theta }}{g} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\\
\,\,\tan \theta = 4\,\,or\,\theta = {\tan ^{ – 1}}\left( 4 \right)
\end{array}$
Standard 11
Physics
