A) \[{{30}^{o}}\]
B) \[{{90}^{o}}\]
C) 45 o
D) \[{{60}^{o}}\]
Correct Answer: D
Solution :
\[\theta ={{\tan }^{-1}}\left( \frac{{{v}^{2}}}{rg} \right)={{\tan }^{-1}}\left[ \frac{{{(14\sqrt{3})}^{2}}}{20\sqrt{3}\times 9.8} \right]={{\tan }^{-1}}[\sqrt{3}]\]\[=60{}^\circ \]
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