Question

A ball is thrown vertically upwards from the top of tower of height `h` with velocity `v` . The ball strikes the ground after time.
A. `v/g[1+sqrt(1+(2gh)/(v^(2)))]`
B. `v/g[1+sqrt(1-(2gh)/(v^(2)))]`
C. `(v)/(g)(1+(2gh)/(v^(2)))^(1//2)`
D. `(v)/(g)(1-(2gh)/(v^(2)))^(1//2)`

Answer 1

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)))]`