A body is thrown vertically upwards from `A`. The top of a tower . It reaches the fround in time `t_(1)`. It it is thrown verically 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.
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Explanation

Suppose the body be projected vertically upwards from `A` with a speed `u_(0)`.
Using equation `s=ut+((1)/(2))a t^(2)`
For first case `-h=u_(0)t_(1)-((1)/(2))g t_(1)^(2)` (i)
For second case, `-h=-u_(0)t_(2)-((1)/(2))g t_(2)^(2)` (ii)
From (i)-(ii) `rArr 0=u_(0)(t_(2)+t_(1))+((1)/(2))g (t_(2)^(2)-t_(12)^(2))`
or `u_(0)=((1)/(2)) g (t_(1)-t_(2))` (iii)
Put the value of `u_(0)` in (ii), we get
`-h=-((1)/(2))g (t_(1)-t_(2))t_(2)-((1)/(2))g t_(2)^(2)`
`rArr h=(1)/(2) g t_(1)t_(2)`
For third case, `u=0, t=`?
`-h=0 xxt-((1)/(2))g t^(2)`
or `h=((1)/(2))g t^(2)`
Combining (iv) and (v), we get
`(1)/(2) g t^(2)=(1)/(2) g t_(1)t_(2)` or `t=sqrtt_(1)t_(2)`.