Question

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

Answer 1

According to conservation of energy.
`mgH=(1)/(2) mv^(2)+mgh_(2)impliesmg(H-h_(2))=(1)/(2)mv^(2)`
where, H=height of the first hill, `h_(1)`=height of the second hill
`h_(2)`= height of the horizontal base
v=velocity attained by the ball
`implies v=sqrt(2g(100-20))=sqrt(2xx10xx80)=40 ms^(-1)`