Rotational Motion Question 32
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^{0} $ with the horizontal. The ball strikes the ground at B. What is the value of the distance AB? (Moment of inertia of spherical shell of mass m and radius R about its diameter $ =\frac{2}{3}{mR^{2}}) $
[JEE ONLINE 22-04-2013]
Options:
A) 1.87 m
B) 2.08 m
C) 1.57 m
D) 1.77 m
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] 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 = $\frac{2}{3})R$
Horizontal range $AB=v^2sin2 \theta $=
$\frac{\sqrt\frac{2gh}{1+\frac{k^2}{R^2}} sin(2×30 \circ)}{g}=2.08m$
