A stone is projected from the ground in such a direction so as to hit a bird on the top of a telegraph post of height h and attains the maximum height of 2h above the ground. If at the insatant of projection, the bird were to fly away horizontally with a uniform speed, find the ratio between the horizontal velocity of bird and the horizontal component of velocity of stone, if the stone hits the bird while descending.

Explanation

`(u^(2) sin^(2) theta)/(2 g) = 2h rArr u sin theta = 2sqrt(g h)`
Let the stone reaches to the bird after time `t`
`y = u sin theta t - (1)/(2) g t^(2)`
`h = u sin theta t - (1)/(2) g t^(2)`
`g t^(2) - 2u sin theta t + 2h = 0`
`t_(1) + t_(2) = ((- 2u sin theta))/(g) = (2 u sin theta)/(g)`
`t_(1) t_(2) = (2 h)/(g)`
`t_(1)` : time taken by the stone from `O` to `A`
`t_(2)` : time taken by the stone from `O` to `B`
`(t_(2) - t_(1))` : time taken by the stone to move from `A` to `B`
`(t_(2) - t_(1))^(2) = (t_(2) + t_(1))^(2) - 4 t_(1)t_(2)`
`= ((4 u^(2) sin^(2) theta))/(g^(2)) - (8h)/(g)`
`= (16 h)/(g) - (8 h)/(g) = (8 h)/(g)`
`t_(2) - t_(1) = 2 sqrt(2)sqrt((h)/(g))` (i)
`t_(2) + t_(1) = (2 u sin theta)/(g) = (2)/(g).2sqrt(g h) = 4 sqrt((h)/(g))` (ii)
Adding (i) and (ii), we get
`t_(2) = (2sqrt(2) + 4) sqrt((h)/(g)) rArr (2 + sqrt(2)) sqrt((h)/(g))`
`t_(2)` : time taken by the bird to move from `A` to `B`
`d = v_(0) t_(2) = u cos theta(t_(2) - t_(1))`
`(v_(0))/(u cos theta) = ((t_(2) - t_(1)))/(t_(2)) = (2 sqrt(2))/(2 + sqrt(2)) = (2)/(sqrt(2) + 1)`