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

Taking dowanwards direction as the positive direction.

`+h=-ut_(1)+(1)/(2)g t_(1)^(2)` …(i)
`+h=ut_(2)+(1)/(2)g t_(2)^(2)`
Multiplying Eq. (i) by `t_(2)` and Eq. (ii) by `t_(!)` and adding, we get
`h(t_(1)+t_(2))=(1)/(2)g t_(1)t_(2)(t_(1)+t_(2))` or `h=(1)/(2)g t_(1)t_(2)`
For free fall from rest, `h=(1)/(2)g t^(2)`
`therefore t^(2)=t_(1)t_(2) rArr t=sqrt(t_(1)t_(2))`