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 \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] \[=\frac{{{u}^{2}}\sin {{90}^{o}}}{g}\] \[R=\frac{{{u}^{2}}}{g}\] \[\therefore \] \[\frac{R}{2}=\frac{{{u}^{2}}}{2g}\] \[\therefore \] \[\tan \alpha =\frac{H}{R/2}\] \[=\frac{{{u}^{2}}/4g}{{{u}^{2}}/2g}\] \[\tan \alpha =\frac{1}{2}\] \[\alpha ={{\tan }^{-1}}\left( \frac{1}{2} \right)\]You need to login to perform this action.
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