Part C) is an entire new question and that is why there are two answers here to the problem.
Taking the positive axis downwards, the distance that the small ball travels is
`s = (v-eu)*t +(g*t^2)/2`
Also this distance for the bucket that has a speed `v` after collision is
By equating the above two expressions one gets
`(v-eu)*t +(g*t^2)/2 =v*t`
`(g*t^2)/2 -eu*t =0`
`t*(g*t/2 -e*u) =0`
This equation has two solutions
`t=0` initial time when the ball bounces first from the bucket
`t = (2eu)/g` time when the ball bounces second from the bucket.