JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Kinematics

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 the angle

4 5^{0}

with respect to the vertical. When the balloon climbs up a future height

h_{2}

, it is seen at the angle

6 0^{0}

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 3 0^{\circ} =

0.5774):

Answer: (D)

Solution

The balloon stays directly above

A

. At first the girl is at

B

, a distance

d

from

A

, and her line of sight makes

4 5^{\circ}

with the vertical, so

tan 4 5^{\circ} = \frac{d}{h _{1}}

and

h_{1} = d

. After the balloon climbs to a height

h_{1} + h_{2}

she has moved to

C

, now

d + 2.464 d = 3.464 d

from

A

, and the sight line makes

6 0^{\circ}

with the vertical, so

tan 6 0^{\circ} = \frac{3.464 d}{h _{1} + h _{2}}

. Since

3.464 \approx 23

and

tan 6 0^{\circ} = 3

h_{1} + h_{2} = \frac{2 3 d}{3} = 2 d

. Subtracting

h_{1} = d

leaves

h_{2} = d

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option