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

A)  \[{{60}^{o}}\]           

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

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

D) \[{{45}^{o}}\] 

Correct Answer: B

Solution :

 Height of projectile \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{u}^{2}}{{\sin }^{2}}{{45}^{o}}}{2g}\] \[\Rightarrow \]\[H=\frac{{{u}^{2}}}{4g}\]                                            ?(i) Range of projectile \[R=\frac{{{u}^{2}}\sin 2\theta }{g}=\frac{{{u}^{2}}\sin {{90}^{o}}}{g}\] \[\Rightarrow \]\[R=\frac{{{u}^{2}}}{g}\] \[\tan \alpha =\frac{H}{R/2}\] In         \[\Delta {\mathrm O}\Alpha \Beta =\frac{{{u}^{2}}/4g}{{{u}^{2}}/2g}\] \[\tan \alpha =\frac{1}{2}\Rightarrow \alpha ={{\tan }^{-1}}\left( \frac{1}{2} \right)\]