Question

3. In the figure shown above, a hot air balloon rising straight up from the ground is tracked by a television
camera 300 ft from the liftoff point. At the moment the camera's elevation angle is a/6, the balloon is
rising at the rate of 80 ft/min . At what rate is the angle of elevation changing at that moment?
(A) 0.12 radian per minute
(B) 0.16 radian per minute
(C) 0.2 radian per minute
(D) 0.4 radian per minute
(E) 0.6 radian per minute

Answer 1

The movement of the balloon is an illustration of angular displacement.

The rate at which the angle of elevation is changing is 0.2 radian per minute

The parameters are given as:

$\theta = \frac{\pi}{6}$

$d = 300ft$

$v = 80ft\ min^{-1}$

See attachment for the balloon's elevation

The angular velocity (w) of an object is calculated as:

$w = \frac vr$

Substitute 80 for v

$w = \frac{80}r$

Considering the attached image

The value of r is calculated from the following cosine ratio

$\cos(\theta) = \frac{300}{r}$

Make r the subject

$r = \frac{300}{\cos(\theta)}$

Substitute $\theta = \frac{\pi}{6}$

$r = \frac{300}{\cos(\frac{\pi}6)}$

Substitute $r = \frac{300}{\cos(\frac{\pi}6)}$ in $w = \frac{80}r$

$w = \frac{80}{\frac{300}{\cos(\frac{\pi}6)}}$

In trigonometry ratios:

$\cos(\frac{\pi}6)} = \frac{\sqrt 3}{2}$

So, we have:

$w = \frac{80}{\frac{300}{\frac{\sqrt 3}2}}$

This gives:

$w = \frac{80}{\frac{600}{\sqrt 3}}$

$w = \frac{80\sqrt 3}{600}$

$w = 0.2309$

Approximate

$w = 0.2$

Hence, the rate at which the angle of elevation changes is $0.2\ rad\ min^{-1}$

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