A stone is projected from a point on the ground so as to hit a bird on the top of a vertical pole of height h and then attain a maximum height 2 h above the ground. If at the instant of projection the bird flies away horizontally with a uniform speed and if the stone hits the bird while descending, then the ratio of the speed of the bird to the horizontal speed of the stone is 

Maximum height of the projectile is given by the expression

${\mathrm{h}}_{\max }=\frac{{\mathrm{v}}_0^2{\sin }^2\theta }{2g}$

$\Rightarrow 2 \mathrm{h}=\frac{{\mathrm{v}}_0^2{\sin }^2\theta }{2 \mathrm{g}}\Rightarrow {\mathrm{v}}_0=\frac{2\sqrt{\mathrm{gh}}}{\sin \theta }$ ……… (i)

$$\begin{gathered} \mathrm{y}={\mathrm{v}}_0\sin \theta \mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2 \\ \Rightarrow \mathrm{h}=\frac{2\sqrt{\mathrm{gh}}}{\sin \theta }\sin \theta \mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2\Rightarrow {\mathrm{gt}}^2-4\sqrt{\mathrm{ght}}+2 \mathrm{h}=0 \\ \mathrm{t}=\frac{4\sqrt{\mathrm{gh}}\pm \sqrt{16\mathrm{gh}-8\mathrm{gh}}}{2 \mathrm{g}}=\frac{4\sqrt{\mathrm{gh}}\pm 2\sqrt{2\mathrm{gh}}}{2 \mathrm{g}} \\ \Rightarrow {\mathrm{t}}_1=\sqrt{\frac{\mathrm{h}}{\mathrm{g}}}(2-2\sqrt{2}) \text{and } {\mathrm{t}}_2=\sqrt{\frac{\mathrm{h}}{\mathrm{g}}}(2+\sqrt{2}) \end{gathered}$$

At these two times, the projectile is at the same height as the bird. 

Let v be the speed of the bird. For bird to be hit, therefore,

$$\begin{gathered} {\mathrm{vt}}_2={\mathrm{v}}_{\mathrm{a}}\cos \theta \left({\mathrm{t}}_2-{\mathrm{t}}_1\right) \\ \Rightarrow \frac{\mathrm{v}}{{\mathrm{v}}_0\cos \theta }=\frac{{\mathrm{t}}_2-{\mathrm{t}}_1}{{\mathrm{t}}_2}=\frac{2}{\sqrt{2}+1} \end{gathered}$$