Correct Answer - A

For `AB " " AB`

`s=ut+1/2 at^(2)`.

`-h=vt-g/2 t^(2)`

`g/2 t^(2)-vt-h=0`

`t=(vpmsqrt(v^(2)+4xxg/2xxh))/(2xxg/2)`

`t=(vpmsqrt(v^(2)+2gh))/g `

`t=v/g[1+sqrt(1+(2gh)/(v^(2)))], t=v/g[1-sqrt(1+(2gh)/(v^(2)))]` [as time cannot be negative so we neglect it we neglect]

`:. t=v/g[1+sqrt(1+(2gh)/(v^(2)))]`

Aliter

Let `t_(1)` be the time taken by ball from top of tower to the highest point then it will taken again `t_(1)` time to return back to the top of tower Let `t_(2)` be the time taken be ball from top of tower of the ground.

For `t_(1)`: From equation

V=u-gt i.e. 0=V-`gt_(1)` or, `t_(1)=V//g`

For `t_(2)`: from equation

`h=ut+1/2 g t^(2) h=V t^(2)+1/2 g t_(2)^(2)`: or , `g t_(2)^(2)+2Vt_(2)-2h=0`, or `t_(2)=(-2v pm sqrt(4V^(2)+8gh))/(2g)`

Taking (+) sign only (as we are interested in time projection i.e. t=0) `t_(2)=(-Vpmsqrt(V^(2)+2gh))/g`

Note that, -ve time indicate time before the projection

Hence, the time after which the ball strikes ground `T=2t_(1)+t_(2) rArr T=(2v)/g+(-V+sqrt(V^(2)+2gh))/g`

`T=(V+sqrt(V^(2)+2gh))/grArr T=V/g[1+sqrt(1+(2gh)/(V^(2)))]`