a. Since the ball moves inside the tube, it can stay at the top of the tube with zero speed.
`implies v_(B)=0`
Applying `COE`
`1/2m(v_(B)^(2)-v_(A)^(2))=mgh=0`
Putting `v_(A)=v_(B)=0` and `h=H-2r`
we obtain `r=H/2`
b. At the lowest positon `C`, let the (maximm) thrust exerted on the ground be `N`.
`N-mg=ma_(r)=(mv_(c)^(2))/rimpliesN=m[v_(c)^(2)/r+g]`
where `v_(c)` can be obtained by coserving energy between `A` and `C` as
`1/2mV_(c)^(2)=mgH=0implies v_(c)^(2)=2gH`
`impliesN=m[(sqrt(2g)^(2))/r+g]`
Putting `r=H/2` we obtain `N=m[(2gH)/(H/2)+g]=5mg`