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?
0.63 meter/sec
2.16 meter/sec
3.87 meter/sec
0.72 meter/sec
Explanation:
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, $\angle$BCA = 60°
tan60°=$\dfrac{BA}{CA}$
=>√3=$\dfrac{BA}{150}$
BA=150√3
i.e, the distance travelled by the balloon =150√3 meters
time taken = 2 min = 2 × 60 = 120 seconds
Speed =$\dfrac{Distance}{Time}$
=$\dfrac{150\sqrt{3}}{120}$=1.25√3
=1.25×1.73=2.16 meter/second
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, $\angle$BCA = 60°
tan60°=$\dfrac{BA}{CA}$
=>√3=$\dfrac{BA}{150}$
BA=150√3
i.e, the distance travelled by the balloon =150√3 meters
time taken = 2 min = 2 × 60 = 120 seconds
Speed =$\dfrac{Distance}{Time}$
=$\dfrac{150\sqrt{3}}{120}$=1.25√3
=1.25×1.73=2.16 meter/second