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. $1.87\,m$
• B. $2.08\,m$
• C. $1.57\,m$
• D. $1.77\,m$

Answer 1

Answer: (A)

$1.87\,m$

Answer 2

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 $AB=\frac{v^{2}\,sin\,2\theta}{g}$
$=\frac{\left(\frac{\sqrt{ 2gh}}{1+k^{2}/R^{2}}\right)sin\left(2\times30^{\degree}\right)}{g}$