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 3 0 ∘ with the horizontal. The ball strikes the ground

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 $3 0^{\circ}$ 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 $= \frac{2}{3} m R^{2}$ )

TestHub · Accepted Answer

Velocity of the tennis ball on the surface of the earth or ground ​ v = 1 + R 2 k 2 ​ 2 gh ​ ​ ( where k = radius of gyration of spherical shell = 3 2 ​ ​ R ) Horizontal range AB = g v 2 sin 2 θ ​ = g 1 + k 2 / R 2 2 gh ​ ​ ) 2 sin ( 2 × 3 0 ∘ ) ​ ​

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