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Question

(B) Let the initial position of the bird be $P$ at the top of the pole $PQ = 20 \ m$. The angle of elevation from point $O$ is $\angle POQ = 45^o$.In

Vedclass · Accepted Answer

$20(\sqrt{3} - 1)$

Vedclass · Answer

(B) Let the initial position of the bird be $P$ at the top of the pole $PQ = 20 \ m$. The angle of elevation from point $O$ is $\angle POQ = 45^o$.In $\Delta POQ$,$	an 45^o = \frac{PQ}{OQ}$ $\Rightarrow 1 = \frac{20}{OQ}$ $\Rightarrow OQ = 20 \ m$.After one second,the bird reaches position $P'$ such that $P'Q' = 20 \ m$ (since it flies horizontally). The angle of elevation from $O$ is $\angle P'OQ' = 30^o$.In $\Delta P'OQ'$,$	an 30^o = \frac{P'Q'}{OQ'}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{20}{OQ'}$ $\Rightarrow OQ' = 20\sqrt{3} \ m$.The distance covered by the bird in one second is $QQ' = OQ' - OQ = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1) \ m$.Since the time taken is $1 \ s$,the speed of the bird is $\frac{20(\sqrt{3} - 1)}{1} = 20(\sqrt{3} - 1) \ m/s$.

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