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JEE Main 202125 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Morning Shift
Subject
Mathematics
Chapter
Heights and Distances

JEE Main 2021 · 25 February · Morning Shift · Mathematics · Heights and Distances

A man is observing, from the top of a tower, a boat speeding...

A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point AA, with uniform speed. At that point, angle of depression of the boat with the man's eye is 30°30 ° (Ignore man's height). After sailing for 2020 seconds, towards the base of the tower (which is at the level of water), the boat has reached a point BB, where the angle of depression is 45°.45 ° . Then the time taken (in seconds) by the boat from BB to reach the base of the tower is :A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A , with uniform

Options

  1. A1010
  2. B10(3−1)10 ( \sqrt{3} - 1 )
  3. C10310\sqrt{3}
  4. D10(3+1)10 ( \sqrt{3} + 1 )

Correct answer

D. 10(3+1)10 ( \sqrt{3} + 1 )

Step-by-step solution

A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A , with uniform

Let speed of boat is um/sum / s and height of tower
is hh meter & distance AB=xAB = x metre

∴x=hcot30°−hcot45°∴ x = hcot30 ° - hcot45 °

⇒x=h(3−1)⇒ x = h ( \sqrt{3} - 1 )

∴u=x20=h(3−1)20m/s∴ u = \dfrac{x}{20} = \dfrac{h ( \sqrt{3} - 1 )}{20}m / s

∴∴ Time taken to travel from BB to CC (Distance
=h= h meter)

=hu=hh(3−1)20=203−1=10(3+1)sec.= \dfrac{h}{u} = \dfrac{h}{h\dfrac{( \sqrt{3} - 1 )}{20}} = \dfrac{20}{\sqrt{3} - 1} = 10 ( \sqrt{3} + 1 ) sec .

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