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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
$\frac{\sqrt{2}}{\sqrt{2}+1}$
$\frac{\sqrt{2}}{\sqrt{2}-1}$
$\frac{1}{\sqrt{2}}+\frac{1}{2}$
$\frac{2}{\sqrt{2}+1}$
Solution
(d)
$2 h=\frac{u_y^2}{2 g}$
or $u_y=2(\sqrt{g} h)$
$\operatorname{Now}\left(t_2-t_1\right) u_x=t_2 v_x$ or $\frac{v_x}{u_x}=\frac{t_2-t_1}{t_2}……..(i)$
Further $h=u_y t-\frac{1}{2}(g t)^2$
or $g t^2-2 u_y t+h=0$
or $g t^2-4(\sqrt{g h}) t+2 h=0$
$t_1= \frac{4 \sqrt{g h}-\sqrt{16 g h-8 g h}}{2 g}=(2-\sqrt{2})\left(\frac{\sqrt{h}}{g}\right)$
Substituting in Eq. (i) we have,
$\frac{v_x}{u_x}=\frac{2}{(\sqrt{2})+1}$
