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. `10sqrt30 m//s`
C. 40 m/s
D. 20 m/s

Answer

C

Explanation

`mg(h_1-h_2) =1/2 mv^2`
`implies v = sqrt(2g(h_1-h_2))=sqrt(2 xx10 xx (100 - 20)) = 40 m//s`