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

Let Vy be the vertical component of the stone's velocity and Vx be it's horizontal velocity.

Max height = 2H = (Vy^2)/2g

Vy^2 = 4gH

Now, the time at which the stone reaches height H. There will be two such times. Let them be T and T', where T > T'.

H = (Vy)t - g(t^2)/2

Solutions of this are T, T'.

T = Vy + sqrt[Vy^2 - 2gH]/g

We need T - T' for the time between the 2 heights.

T - T' = 2sqrt{Vy^2 - 2gH}/g

Finally, (velocity of bird)/(Vx) = (T - T')/T = 2/[(sqrt2) + 1]