A stone is dropped from a height h. Simultaneously, another stone is thrown up from the ground which reaches a height 4 h. The two stones cross each other after time

Options

A. `sqrt( (h)/(8g))`
B. `sqrt(8gh)`
C. `sqrt(2gh)`
D. `sqrt((h)/(2g))`

Answer

A

Explanation

For first stone u = 0
For second stone `(u^2)/(2g) = 4h rArr u^2 = 8gh`

` therefore u = sqrt(8gh)`
Now `h_1 = 1/2 g t^2`
`h_2 = sqrt(8ght) - 1/2 g t^2`
where, t =time to cross each other.
`because h_1 + h_2 = h`
`rArr 1/2 g t^2 + sqrt(8ght) - 1/2 g t^2 = h rArr t = (h)/(sqrt(8gh)) = sqrt( (h)/(8g))`