A body is thrown vertically upwards from `A`. The top of a tower . It reaches the ground in time `t_(1)`. It it is thrown vertically downwards from `A` with the same speed it reaches the ground in time `t_(2)`, If it is allowed to fall freely from `A`. then the time it takes to reach the ground.
.

Options

A. `t=(t_(1)+t_(2))/(2)`
B. `t=(t_(1)+t_(2))/(2)`
C. `t=sqrt(t_(1)t_(2))`
D. `t=sqrt((t_(1))/(t_(2)))`

Answer

C

Explanation

Let downwards direction is +ve
`h=-ut_(1)+(1)/(2)"gt"_(1)^(2)`.....(1)
`h=ut_(2)+(1)/(2)"gt"_(2)^(2)` ......(2)
Multiplying eq^(n). (1) by `t_(2)` and `eq^(n)`. (2) by `t_(1)`
`ht_(2)=-ut_(1)t_(2)+(1)/(2)"gt"_(1)^(2)t_(2)`....(3)
`ht_(1)=ut_(2)t_(1)+(1)/(2)"gt"_(2)^(2)t_(1)` ....(4)
adding `eq^(n).` (3) `&` (4)
`ht_(1)+ht_(2) =(1)/(2) "gt"_(1)^(2)t_(2)+(1)/(2)"gt"_(2)^(2)t_(1)`
`rArrh(t_(1)+t_(2))=(1)/(2)"gt"_(1)t_(2)`
`rArr h=(1)/(2)"gt"_(1)t_(2)` ......(5)
for free fall when u=0, then
`h=(1)/(2) "gt"^(2)`.....(6)
From `eq^(n), (5) & (6)`
`(1)/(2)"gt"^(2)=(1)//(2)"gt"_(1)t_(2)rArrt=sqrt(t_(1)t_(2))`