A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O…

Question

A bird is sitting on the top of a vertical pole $20$ m high and its elevation from a point $O$ on the ground is $${45^ \circ }$$. It files off horizontally straight away from the point $O$. After one second, the elevation of the bird from $O$ is reduced to $${30^ \circ }$$. Then the speed (in m/s) of the bird is :

Accepted Answer

Let the speed be $y$ $m/sec$.

Let $AC$ be the vertical pole of height $20$ $m.$

Let $O$ be the point on the ground such that $$\angle AOC = {45^ \circ }$$

Let $OC = x$

Time $t=1$ $s$

From $$\Delta AOC,\,\,\tan {45^ \circ } = {{20} \over x}\,\,\,\,\,\,\,.....\left( i \right)$$

and from  $$\Delta BOD,\,\,\tan {30^ \circ } = {{20} \over {x + y}}...\left( {ii} \right)$$

From $(i)$ and $(ii),$ we have $x=20$

and $${1 \over {\sqrt 3 }} = {{20} \over {x + y}}$$

$$ \Rightarrow {1 \over {\sqrt 3 }} = {{20} \over {20 + y}}$$

$$ \Rightarrow 20 + y = 20\sqrt 3 $$

So, $$y = 20\left( {\sqrt 3  - 1} \right)\,\,i.e.,$$

speed $$ = 20\left( {\sqrt 3  - 1} \right)m/s$$

Solutions · 1