Lucky Bird Hunter

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]

Sorry for non-formatting, I am unable to do so properly.

The bird would be at a height $h$ twice during it's motion. The total time from the point of projection to the point where stone and bird collide can be given by the bigger root of $h= v_{0} \sin{\alpha} t_{AB} - \frac{1}{2} g {t_{AB}}^{2}$ Now, we know that $v_{0} \sin{\alpha} = 2 \sqrt{gh}$ by writing its maximum height. Hence, substituting this in the first equation and solving quadratic gives us $t_{AB} = \sqrt{\frac{2h}{g}} (\sqrt{2} \pm 1)$ . Of course, the total time from the point of projection to the point where stone and bird collide would be $T= \boxed{\sqrt{\frac{2h}{g}} (\sqrt{2} + 1)}$ . Since the bird is moving with a constant velocity (Let it be $v_{b}$ ), we can write: $\frac{R_{BD}}{v_{b}} = T$ , where $R_{BD}$ is the distance the bird travels. This can of course be written as $2v_{0} \cos{\alpha} \cdot \sqrt{\frac{2h}{g}}$ , considering it to be the range of a smaller projectile thrown with vertical velocity $\sqrt{2gh}$ and horizontal velocity $v_{0} \cos{\alpha}$ . Now, using $\frac{R_{BD}}{v_{b}} = T$ , we get

${\boxed{\boxed{\frac{v_{0} \cos{\alpha}}{v_{b}} = \frac{2}{\sqrt{2} +1}}}}$