The angle of elevation of a hot-air balloon climbing vertically changes from 30° at 10:00 AM to 60° at 10:02 AM as observed from a point 300 m away from the take-off point. What is the upward speed of the balloon?

Correct option is D

Let the balloon take off at the point B and at 10 am balloon is at the point E and 10:02 am, balloon is at point A.
BC = 300 m
In ΔEBC
tan 30 = EB/BC
=> 1/√3 = EB/300
=> EB = 300/√3
In ΔABC
tan 60 = AB/BC
=> √3 = AB/300
AB = 300 √3
AE = AB - EB
AE = 300√3 - 300/√3
AE = [900 - 300]/√3
AE = [600/√3] × [√3/√3]
AE = 600√3/3 = 200√3
Distance covered by balloon in 2 min = 200 √3
Distance covered by balloon in 1 min = 100 √3 = 100 × 1.732 = 173.2
Speed of the balloon = 173.2 m/min
Speed of the balloon = 173.2/60 m/s = 2.9 m/sec
Short Trick:
√3 unit = 300
=> 2 unit = [300/√3] × 2 = 600/√3
Distance covered by balloon in 2 min = 600/√3
∴ Speed of balloon = [600/√3] × [1/(2 × 60)] = 2.9 m/sec