JEE PYQ: Mathematics In Physics Question 11
Question 11 - 2020 (05 Sep Shift 1)
A balloon is moving up in air vertically above a point A on the ground. When it is at a height $h_1$, a girl standing at a distance $d$ (point B) from A (see figure) sees it at an angle 45° with respect to the vertical. When the balloon climbs up a further height $h_2$, it is seen at an angle 60° with respect to the vertical if the girl moves further by a distance 2.464 $d$ (point C). Then the height $h_2$ is (given $\tan 30° = 0.5774$):
(1) 1.464 $d$
(2) 0.732 $d$
(3) 0.464 $d$
(4) $d$
Show Answer
Answer: (4)
Solution
From figure/trigonometry, $\frac{h_1}{d} = \tan 45°$ $\therefore h_1 = d$ And, $\frac{h_1 + h_2}{d + 2.464d} = \tan 30°$ $\Rightarrow (h_1 + h_2) \times \sqrt{3} = 3.46d$ $\Rightarrow (h_1 + h_2) = \frac{3.46d}{\sqrt{3}} \Rightarrow d + h_2 = \frac{3.46d}{\sqrt{3}}$ $\therefore h_2 = d$
BETA
