A balloon leaves the earth at a point A and rises vertically at uniform speed. At the end of 2 minutes, John finds the angular elevation of the balloon as 60°. If the point at which John is standing is 150 m away from point A, what is the speed of the balloon?

Solution(By Examveda Team)

Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.
Given that CA = 150 m, ∠BCA = 60°
$$\eqalign{ & \tan {60^ \circ } = \frac{{BA}}{{CA}} \cr & \sqrt 3 = \frac{{BA}}{{150}} \cr & BA = 150\sqrt 3 \cr} $$
i.e, the distance travelled by the balloon = $$150\sqrt 3 $$   meters
time taken = 2 min = 2 × 60 = 120 seconds
$$\eqalign{ & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{150\sqrt 3 }}{{120}} = 1.25\sqrt 3 \cr} $$
= 1.25 × 1.73 = 2.16 meter/second