A) 1.87 m
B) 2.08 m
C) 1.57 m
D) 1.77 m
Correct Answer: B
Solution :
Velocity of the tennis ball on the surface of the earth or ground \[v=\sqrt{\frac{2gh}{1+\frac{{{k}^{2}}}{{{R}^{2}}}}}\] (where k = radius of gyration of spherical shell\[=\sqrt{\frac{2}{3}}R\]) Horizontal range \[AB=\frac{{{v}^{2}}\sin 2\theta }{g}\] \[=\frac{{{\left( \sqrt{\frac{2gh}{1+{{k}^{2}}/{{R}^{2}}}} \right)}^{2}}\sin (2\times {{30}^{o}})}{g}\]You need to login to perform this action.
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