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