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((1_(1)/(t_(2))`

Answer

C

Explanation

Suppose the body be projected vertically upwards from `A` with s speed `u_(0)`,
`Uisng equation `s=ut + ((1)/(2)) at^(2)`, we get
For first case: `-h=u_(0)t_(10 - ((1)/(2)) g t_(1)^(2)`
For second acse:`-h =- u_(0)t_(2) -((1)/(2)) g t_(2)^(2)`
(i) -(ii) `rArr 0=u_(0) (t_(2) +t_(1)) + ((1)/(2)) g (t_(2)^(2)-t_(1)^(2))`
`rArr u_(0) =((1)/(2))g (t_(1)-t_(2))`
Putting the value of `u_(0)` in (ii), we get
`-h=- ((1)/(2)) g (t_(1)-t_(2)) -((1)/(2)) g t_(2)^(2)`
`rArr h=(1)/(2) g t_(1)t_(2)` (iv)
For thir case: `u =0, t=`?
`-h=0xx t-((1)/(2 0 g t^(2)` or `h=((1)/(2)) g t^(2)` (v)
Combining Eq. (iv) and Eq. (v), we get
`(1)/(2) g t^(2) =(1)/(2) g t_(1))t_(2)` or `tsqrt t_(1) t_(2)`.