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 $30 °$ 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 = $\frac{2}{3} m R^{2}$ )

Question

Smart Achievers · Accepted Answer

(1)

Velocity of the tennis ball on the surface the earth or ground

$m$ (where $k =$ radius of gyration of spherical shell $= \sqrt{\frac{2}{3}} R$

Horizontal range AB $= \frac{v^{2} \sin 2 θ}{g}$ $= \frac{{(\sqrt{\frac{2 g h}{1 + \frac{k^{2}}{R^{2}}}})}^{2} \sin (2 \times 30 °)}{g} = 1.87 m$

Solution

1 1.87 m

2 2.08 m

3 1.57 m

4 1.77 m

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