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?

Answer: Option B

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°
$$\begin{array}{rl}& \tan 60^{\circ }=\frac{BA}{CA}\\ & \sqrt{3}=\frac{BA}{150}\\ & BA=150\sqrt{3}\end{array}$$
i.e, the distance travelled by the balloon = $$150\sqrt{3}$$   meters
time taken = 2 min = 2 × 60 = 120 seconds
$$\begin{array}{rl}& \text{Speed}=\frac{\text{Distance}}{\text{Time}}\\ & =\frac{150\sqrt{3}}{120}=1.25\sqrt{3}\end{array}$$
= 1.25 × 1.73 = 2.16 meter/second