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Question:
A tennis ball (treated as hollow spherical shell) starting from $O$ rolls down a hill. At point A the ball becomes air borne leaving at an angle of $30^\circ$ with the horizontal. The ball strikes the ground at $B$. What is the value of the distance $AB$ ?
(Moment of inertia of a spherical shell of mass $m$ and radius $R$ about its diameter $=\frac{2}{3}mR^{2}$ )
A tennis ball (treated as hollow spherical shell) starting from $O$ rolls down a hill. At point A the ball becomes air borne leaving at an angle of $30^\circ$ with the horizontal. The ball strikes the ground at $B$. What is the value of the distance $AB$ ?
(Moment of inertia of a spherical shell of mass $m$ and radius $R$ about its diameter $=\frac{2}{3}mR^{2}$ )
Updated On: Nov 10, 2024
- $1.87\,m$
- $2.08\,m$
- $1.57\,m$
- $1.77\,m$
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The Correct Option is A
Solution and Explanation
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 \(AB=\frac{v^{2}\,sin\,2\theta}{g}\)
\(=\frac{\left(\frac{\sqrt{ 2gh}}{1+k^{2}/R^{2}}\right)sin\left(2\times30^{\degree}\right)}{g}\)
\(v=\sqrt{\frac{2gh}{1+\frac{k^{2}}{R^{2}}}}\) ( where k = radius of gyration of spherical shell \(=\sqrt{\frac{2}{3}R}\) )
Horizontal range \(AB=\frac{v^{2}\,sin\,2\theta}{g}\)
\(=\frac{\left(\frac{\sqrt{ 2gh}}{1+k^{2}/R^{2}}\right)sin\left(2\times30^{\degree}\right)}{g}\)
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