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 \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{u}^{2}}{{\sin }^{2}}{{45}^{o}}}{2g}\]You need to login to perform this action.
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