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JEE Main 20146 April
Exam
JEE Main
Year
2014
Date
6 April
Subject
Mathematics
Chapter
Heights and Distances

JEE Main 2014 · 6 April · Mathematics · Heights and Distances

A bird is sitting on the top of a vertical pole 20m high and its...

A bird is sitting on the top of a vertical pole 20m20m high and its elevation from a point OO on the ground is 45°45 °. It flies off horizontally straight away from the point OO. After one second, the elevation of the bird from OO is reduced to 30°30 °. Then the speed (in m/sm / s) of the bird isA bird is sitting on the top of a vertical pole 20m high and its elevation from a point O on the ground is 45 ° . It

Options

  1. A2022 0 \sqrt{2}
  2. B20(3−1)2 0 \left(\sqrt{3} - 1\right)
  3. C40(2−1)4 0 \left(\sqrt{2} - 1\right)
  4. D40(3−2)4 0 \left(\sqrt{3} - \sqrt{2}\right)

Correct answer

B. 20(3−1)2 0 \left(\sqrt{3} - 1\right)

Step-by-step solution

A bird is sitting on the top of a vertical pole 20m high and its elevation from a point O on the ground is 45 ° . It

AD=20mAD = 20m

AO=20tan⁡45∘=201=20mAO = \dfrac{20}{\tan 4{5}^{∘}} = \dfrac{20}{1} = 20m

CB=20mCB = 20m

∴BO=20tan30∘=20(13)=203m∴ BO = \dfrac{20}{tan{30}^{∘}} = \dfrac{20}{\left(\dfrac{1}{\sqrt{3}}\right)} = 20\sqrt{3}m

∴AB=BO−AO∴ AB = BO - AO

=203−20= 2 0 \sqrt{3} - 2 0

=20(3−1)m= 20\left(\sqrt{3} - 1\right)m

It took 1sec1sec to travel distance DCDC.

∴∴ Speed =20(3−1)= 20\left(\sqrt{3} - 1\right)m / sec.

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