Question

A balloon rises at the rate of 10 ft/sec from a point on the ground 100 feet from an observer. Find the rate of change of the angle of elevation of the balloon from the observer when the balloon is 100 feet above the ground.

Answer 1

The rate of change of the angle of elevation of the balloon from the observer when the balloon is 100 feet above the ground is  1/20 degree/sec.

What is angle of elevation?

It is described as a relationship between an oblique line from the observer's eye to an item above his eye and the horizontal plane. This angle eventually develops above the surface. The angle of elevation is made in such a way that it is above the observer's eye, as the name suggests.

Calculation for the rate of change of angle of elevation-

According to the question,

A balloon rises at the rate of 10 ft/sec from a point on the ground 100 feet from an observer.

The balloon is 100 feet above the ground.

Let the height of the balloon be 'h' ft.

Then tanØ = h/100

h = 100tanØ

Differentiate the above equation with respect to time.

dh/dt = 100 sec² Ø (dØ/dt)

We know dh/dt = 10 ft/sec.

when h = 100

tanØ = 100/100 = 1/1

tanØ = Perpendicular/Base

On comparing above two equations,

Perpendicular = 1; Base = 1.

By Pythagoras the hypotenuse is,

H² = P² + B²

    = 1² + 1²

    = 2

H = √2

Now, calculate the value of cosØ.

cosØ = B/H = 1/√2

Squaring both side

cos² Ø = 1/2

sec² Ø = 1/cos² Ø

sec² Ø = 2

The main equation was,

dh/dt = 100 sec² Ø (dØ/dt)

Substitute the values,

dØ/dt = (dh/dt )/100 sec² Ø

         = (10)/(100×2)

dØ/dt = 1/20 degree/sec.

Therefore, the rate of change of the angle of elevation of the balloon from the observer is  1/20 degree/sec.

To know more about the angle of elevation, here

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