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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]
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