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 \[\Delta {\mathrm O}\Alpha \Beta =\frac{{{u}^{2}}/4g}{{{u}^{2}}/2g}\] \[\tan \alpha =\frac{1}{2}\Rightarrow \alpha ={{\tan }^{-1}}\left( \frac{1}{2} \right)\]You need to login to perform this action.
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