question_answer

A projectile is fired at an angle of \[{{45}^{o}}\] with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is                                   [AIPMT (M) 2011]

A) \[{{60}^{o}}\]

B) \[{{\tan }^{-1}}\left( \frac{1}{2} \right)\]

C) \[{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]

D) \[\text{4}{{\text{5}}^{\text{o}}}\]

Correct Answer: B

Solution :

Height of projectile

\[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]

\[H=\frac{{{u}^{2}}{{\sin }^{2}}{{45}^{o}}}{2g}\]

\[H=\frac{{{u}^{2}}}{4g}\]

Range of projectile

\[R=\frac{{{u}^{2}}\sin 2\theta }{g}\]

\[=\frac{{{u}^{2}}\sin {{90}^{o}}}{g}\]

\[R=\frac{{{u}^{2}}}{g}\]

\[\therefore \]            \[\frac{R}{2}=\frac{{{u}^{2}}}{2g}\]

\[\therefore \]      \[\tan \alpha =\frac{H}{R/2}\]

\[=\frac{{{u}^{2}}/4g}{{{u}^{2}}/2g}\]

\[\tan \alpha =\frac{1}{2}\]

\[\alpha ={{\tan }^{-1}}\left( \frac{1}{2} \right)\]