A billiard ball, initially at rest, is given a sharp impulse by a cue. The cue is held horizontally a distance `h` above the centre line as shown in figure. The ball leaves the cue with a speed `v_(0)` and because of its backward slipping eventually acquires a final
speed `(9)/(7)v_(0)` show that `h=(4)/(5)R`
Where `R` is the radius of the ball.

Explanation

If we represent the impulse given to the ball by `J`, then the initial velocity is

`upsilon_(cm) = upsilon_(0) = (J)/(M)` ..(i)
and `omega = (L)/(I) = (M upsilon_(0) xx h)/((2)/(5)M R^(2)) = (5 upsilon_(0)h)/(2R^(2))` ..(ii)
After the application o fimpulse, angular momentum of the ball about `O` will be conserved.
i.e., `L_("initial") = L_("final")`
`I_(cm) omega + M upsilon_(0) xx R = I_(cm) (upsilon)/(R ) + M upsilon xx R`
As `upsilon = (9)/(7) upsilon_(0)`
`:. (2)/(5)M R^(2)((5upsilon_(0)h)/(2R^(2))) + M upsilon_(o)R`
`= ((2)/(5) (MR^(2))/(R ) + MR)(9)/(7) upsilon_(0)`
`M h + MR = (9)/(5)MR`
`Mh = (9)/(5)M R - M R = (4)/(5)MR`
`h = (4)/(5)R`