Q 11-03-150JEE MainJEE Main 2020 (5 Sep, Shift 1)Medium
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 $45^\circ$ with respect to the vertical. When the balloon climbs up a further height $h_2$, it is seen at an angle $60^\circ$ 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 $\tan30^\circ = 0.5774$):
Answer: (D) $d$
At $B$: $\tan45^\circ = \dfrac{d}{h_1} \Rightarrow h_1 = d$.
At $C$ the horizontal distance is $d + 2.464d = 3.464d$:
$$\tan60^\circ = \frac{3.464d}{h_1 + h_2} \Rightarrow h_1 + h_2 = \frac{3.464d}{1.732} = 2d$$
So $h_2 = d$.
Solution by Sreeraj P, M.Sc Physics