A one kg rolling on a smooth horizontal surface at `20 m s^(-1)` comes to the bottom of an inclined plane making an angle of `30^(@)` with the horizontal. Calculate `K.E.` of the ball when it is at the bottom of incline. How far up the incline will the ball roll ? Neglect friction.

Explanation

As is know from theory,
Total `KE` of the ball
`= (7)/(10)m upsilon^(2) = (7)/(10) xx 1 xx(20)^(2) = 280 J`
As work done in stopping the ball `= K.E.` of ball
`:. (mg sin theta) xx s = 280`
`s = (280)/(mg sin theta) = (280)/(1 xx 9.8 xx (1)/(2)) = 57.14 m`