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(D) Let the initial vertical velocity of the stone be $u_y$. The maximum height reached is $H = \frac{u_y^2}{2g} = 2h$. Thus,$u_y = 2\sqrt{gh}$.The ve

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$\frac{2}{\sqrt{2}+1}$

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(D) Let the initial vertical velocity of the stone be $u_y$. The maximum height reached is $H = \frac{u_y^2}{2g} = 2h$. Thus,$u_y = 2\sqrt{gh}$.The vertical position of the stone at time $t$ is $y(t) = u_y t - \frac{1}{2}gt^2$. Setting $y(t) = h$,we get $h = 2\sqrt{gh} t - \frac{1}{2}gt^2$,which simplifies to $gt^2 - 4\sqrt{gh}t + 2h = 0$.Solving for $t$ using the quadratic formula: $t = \frac{4\sqrt{gh} \pm \sqrt{16gh - 8gh}}{2g} = \frac{4\sqrt{gh} \pm 2\sqrt{gh}}{2g}$.The two times are $t_1 = (2-\sqrt{2})\sqrt{\frac{h}{g}}$ (ascending) and $t_2 = (2+\sqrt{2})\sqrt{\frac{h}{g}}$ (descending).Since the stone hits the bird while descending,the time of flight is $t_2$. The horizontal distance covered by the stone is $x = u_x t_2$. The bird starts at distance $x$ from the projection point and moves with speed $v_b$. The distance covered by the bird in time $t_2$ is $d = v_b t_2$. The stone hits the bird at the pole,so the bird must have moved a distance $x$ from the pole in time $t_2$. Thus,$v_b t_2 = u_x t_2 - x_{pole}$,where $x_{pole}$ is the horizontal distance to the pole. Actually,the bird is at the pole at $t=0$ and moves away. The stone reaches the pole at $t_2$. The bird's position at $t_2$ is $x_b = v_b t_2$. The stone's horizontal position at $t_2$ is $x_s = u_x t_2$. Since they collide,$u_x t_2 = x_{pole} + v_b t_2$. Given the stone hits the bird at the pole,$x_{pole} = u_x t_1$. Thus $u_x t_2 = u_x t_1 + v_b t_2$,which gives $\frac{v_b}{u_x} = \frac{t_2 - t_1}{t_2} = \frac{2\sqrt{2}\sqrt{h/g}}{(2+\sqrt{2})\sqrt{h/g}} = \frac{2\sqrt{2}}{2+\sqrt{2}} = \frac{2}{\sqrt{2}+1}$.

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