Rotational Motion

JEE Main level, single correct, medium

185 of 280

Question 185 of 280

A tennis ball (treated as hollow spherical shell) starting from O\mathrm{O} rolls down a hill. At point A\mathrm{A} the ball becomes air borne leaving at an angle of 30∘30^{\circ} with the horizontal. The ball strikes the ground at B\mathrm{B}. What is the value of the distance AB\mathrm{AB} ?
(Moment of inertia of a spherical shell of mass mm and radius RR about its diameter =23mR2=\frac{2}{3} m R^2 )

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Answer: (B) 2.08 m2.08 \mathrm{~m}

Velocity of the tennis ball on the surface of the earth or ground
v=2gh1+k2R2( where k= radius of gyration  of spherical shell =23R) Horizontal range AB=v2sin⁡2θg=2gh1+k2/R2)2sin⁡(2×30∘)g\begin{aligned} & \mathrm{v}=\sqrt{\frac{2 \mathrm{gh}}{1+\frac{\mathrm{k}^2}{\mathrm{R}^2}}}(\text { where } \mathrm{k}=\text { radius of gyration } \\ & \text { of spherical shell } \left.=\sqrt{\frac{2}{3}} \mathrm{R}\right) \\ & \text { Horizontal range } \mathrm{AB}=\frac{\mathrm{v}^2 \sin 2 \theta}{\mathrm{g}} \\ & =\frac{\left.\sqrt{\frac{2 \mathrm{gh}}{1+\mathrm{k}^2 / \mathrm{R}^2}}\right)^2 \sin \left(2 \times 30^{\circ}\right)}{\mathrm{g}} \end{aligned}

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