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 forward english (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

let `omega_(0)` be the angular speed of the ball just after it leaves the cue. The maximum friction acts in forward direction till the slipping continues. Let `v` be the linear speed and `omega` the angular speed when slipping is ceased.
`thereforev=Romega` or `omega=(v)/(R)`
given `v=(9)/(7)v_(0)` ...(i)
`thereforeomega=(9)/(7)(v_(0))/(R)` ...(ii)
Applyingg linear impulse `=` change in angular momentum
`thereforeFdt=mv_(0)` ..(iii)
Angular impulse `=` change in angular momentum
`thereforetaudt=Iomega_(0)` or `(Fh)dt=(2)/(5)mR^(2)omega_(0)` ..(iv)
During the slip angular momentum about bottommost point will remain conserved
i.e., `L_(i)=L_(f)`
or `Iomega_(0)+mRv_(0)=Iomega+mRv`
`therefore(2)/(5)mR^(2)omega_(0)+mRv_(0)=(2)/(5)mR^(2)((9)/(7)(v_(0))/(R))+(9)/(7)mRv_(0)` ..(v)
solving eqs. (iii). (iv) and (v) we get `h=(4)/(5)R`