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

Let the stone is projected with a velocity u at an angle `theta` with the horizontal, we have
`(2h)=((usin theta)^(2))/(2g)` or `u sin theta=2sqrt(gh)`
Suppose t is the time taken by the stone to reach the height h above the ground. Then
`h=u sin theta t-1/2"gt"^(2)` (or) `("gt"^(2))/2- u sin theta+h=0`
As we have `u sin theta =2sqrt(gh)" " :.("gt"^(2))/2-2sqrt(gh)t+h=0`

Solving above equation for t we get
`t=(2sqrt(th)+-sqrt((2sqrt(gh))^(2)-4xxg/2h))/(2xxg/2)=(2sqrt(gh)+-sqrt(2gh))/g`
Let `t_(1)=sqrt((2h)/g)(sqrt(2)-1)` and `t_(2)=sqrt((2h)/g)(sqrt(2)+1)`
Where `t_(1)` and `t_(2)` correspond to P and Q in the figure. Suppose v is the horizontal velocity of the bird. Then `PQ=vt_(2)`
For the motion of stone `PQ=u cos theta (t_(2)-t_(1))" ":.vt_(2)=ucos theta(t_(2)-t_(1))`
which gives `v/(u cos theta)=((t_(2)-t_(1))/(t_(2)))=(sqrt((2gh)/g)(sqrt(2)+1)-(sqrt(2)-1))/(sqrt((2h)/g)(sqrt(2)+1))=2/(sqrt(2)+1)`