A) \[30{}^\circ \]
B) \[45{}^\circ \]
C) \[10.5{}^\circ \]
D) \[26.5{}^\circ \]
Correct Answer: D
Solution :
Angle of incline \[\tan \theta =\frac{{{v}^{2}}}{rg}\] where v is velocity, r the radius of circular path and g the acceleration due to gravity. Given, \[v=9.8\text{ }m/s,\text{ }r=19.6\text{ }m,\] \[g=9.8\text{ }m/{{s}^{2}}\] \[\therefore \] \[\tan \theta =\frac{9.8\times 9.8}{19.6\times 9.8}=\frac{1}{2}\] \[\Rightarrow \] \[\theta ={{26.5}^{o}}\]
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