question_answer

A projectile is fired at an angle of \[45{}^\circ \] with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is                                                                  

A)  \[\text{6}{{\text{0}}^{\text{o}}}\]                          

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                 \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\]                 \[=\frac{{{u}^{2}}\sin {{90}^{o}}}{g}\]                 \[R=\frac{{{u}^{2}}}{g}\] \[\therefore \]  \[\frac{R}{2}=\frac{{{u}^{2}}}{2g}\] \[\therefore \]   \[\tan \alpha =\frac{H}{R/2}\]                   \[=\frac{{{u}^{2}}/4g}{{{u}^{2}}/2g}\]      \[\tan \alpha =\frac{1}{2}\] \[\alpha ={{\tan }^{-1}}\left( \frac{1}{2} \right)\]