A spherical ball of mass `20kg` is stationary at the top of a hill of height `100m` , it rolls down a smooth surface to the ground , then climbs up another bill of height of `30 m` and final rolls down to a horizontal base at a height of `20 m` about the ground . The velocity attained by the ball is

Options

A. `10(m)/(s)`
B. `10sqrt(30)(m)/(s)`
C. `40(m)/(s)`
D. `20(m)/(s)`

Answer

C

Explanation

Ball starts from the top of a hill which is 100 m high and finally rolls down to a horizontal base which is 20 m above the ground so from the conservation of energy `mg(h_1-h_2)=(1)/(2)mv^2`
`impliesv=sqrt(2g(h_1-h_2))=sqrt(2xx10xx(100-20))`
`sqrt(1600)=40(m)/(2)`