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 = $$\frac{2}{3}mR^2$$)

Total energy at O is potential, and at A it is a mix of translational and rotational kinetic energy:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{3}mR^2\right)\left(\frac{v}{R}\right)^2$$

$$mgh = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2$$

$$v^2 = \frac{6gh}{5} = \frac{6 \times 10 \times 1.8}{5} = \frac{108}{5} = 21.6$$

$$v = \sqrt{21.6} \approx 4.65\text{ m/s}$$

$$R_{range} = \frac{v^2 \sin(2\theta)}{g}$$

$$AB = \frac{21.6 \times \sin(60^\circ)}{10}$$

$$AB = 2.16 \times 0.866 \approx 1.87\text{ m}$$