JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 17)
A balloon is moving up in air vertically above a
point A on the ground. When it is at a height h1,
a girl standing at a distanced (point B) from A
(see figure) sees it at an angle 45o with respect
to the vertical. When the balloon climbs up a
further height h2, it is seen at an angle 60o with
respect to the vertical if the girl moves further
by a distance 2.464 d(point C). Then the height
h2 is (given tan 30o = 0.5774)
_5th_September_Morning_Slot_en_17_1.png)
0.464d
d
0.732d
1.464d
Explanation
_5th_September_Morning_Slot_en_17_2.png)
From $$\Delta ABD$$
$$\tan 45 = {{{h_1}} \over d}$$
$$ \Rightarrow 1 = {{{h_1}} \over d} $$
$$\Rightarrow {h_1} = d$$
From $$\Delta ACE$$
$$\tan 30 = {{{h_1} + {h_2}} \over {d + 2.464d}}$$
$$ \Rightarrow $$ $$0.5774 = {{d + {h_2}} \over {3.464d}}$$
$$ \Rightarrow $$ $$d + {h_2} = 0.5774 \times 3.464 \times d$$
$$ \Rightarrow $$ $${h_2} = 2.0001136d - d$$
$$ \Rightarrow $$ $${h_2} = 2.000d - d = d$$
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