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. `sqrt(t_1//t_2)`
B. `sqrt(t_2//t_1)`
C. `sqrt(t_1 t_2)`
D. `t_1 t_2`

Answer

C

Explanation

For downward motion,
`h = ut_1 + (1)/(2) g t_1^2` …(i)
For upward motion,
`h = - ut_2 + (1)/(2) g t_2^2` …(ii)
From (i) & (ii)
`ut_1 + (1)/(2) g t_1^2 = -ut_2 + (1)/(2) g t_2^2`
`rArr u(t_1 + t_2) = (1)/(2) g(t_2^2 - t_1^2)`
:. `h = (g)/(2)(t_1 - t_1) t_1 + (g)/(2) g t_1^2 = (g t_2 t_1)/(2)`
:. Required time `t = sqrt((2h)/(g)) = sqrt(t_1 t_2)`.