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

Answer

D

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(2 xx10 xx (100 -20)) = sqrt(1600) = 40 m//s`.