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

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)`
`rArr v = sqrt(2g(h_(1)-h_(2)))=sqrt(2xx10xx(100-20))`
`= sqrt(1600)=40 m//s`.