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 =(2/3)mR2 )

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° 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 =(2/3)mR2 ) (A) 1.87 m (B) 2.08 m (C) 1.57 m (D) 1.77 m

Tardigrade Admin · Accepted Answer

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

v = \frac{2 g h}{1 + \frac{k ^{2}}{R ^{2}}}

( where k = radius of gyration of spherical shell

= \frac{2}{3} R

)
Horizontal range

A B = \frac{v ^{2} s in 2 θ}{g}

= \frac{( \frac{2 g h}{1 + k ^{2} / R ^{2}} ) s in ( 2 \times 3 0 ^{°} )}{g}

$1.87 m$

$2.08 m$

$1.57 m$

$1.77 m$

1.87m

2.08m

1.57m

1.77m

Velocity of the tennis ball on the surface of the earth or ground v=1+R2k2​2gh​​ ( where k = radius of gyration of spherical shell =32​R​ ) Horizontal range AB=gv2sin2θ​ =g(1+k2/R22gh​​)sin(2×30°)​

$1.87 m$

$2.08 m$

$1.57 m$

$1.77 m$