QJEEM14Mafac9: JEE Main 2014 Mathematics - Class 12 JEEM14Mafac9 Solution: A bird is sitting on the top of a vertical pole m high and its…

Question

Verified Solution • 2025–26 Session moderate

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

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