Question

A missile is fired vertically from a point that is 5 miles from a tracking station and at the same elevation. If the angle of elevation of the missile changes at a constant rate of 2 degrees per second, find the velocity of the missile when the angle of elevation from the tracking station is 30 degrees.

Answer 1

Answer:

the velocity of the missile when the angle of elevation from the tracking station is 30 degrees = $\frac{2 \pi}{27} \ mi/s$

Explanation:

The method employed in solving this question is to relate it to a right- angled triangle;

Now ; if we consider the missle fired vertically from a point  5 miles from tracking station ; with an angle θ and h becoming the height :

Then ;

$tan \ \theta = \frac{h}{5}\\\\h = 5 tan \ \theta$

Differentiating the above equation ; we have

$\frac{dh}{dt} = 5 \ sec^2 \ \theta \ \frac{d \theta}{dt}$

Replacing $v \ with \ \frac{dh}{dt}$ ; $\theta \ with \ 30^0$ and $\frac{d\theta}{dt} \ with \ \ \frac{2 \pi}{180}$; we have :

$v = 5 \ sec^2 \ (30^0) \ (\frac{2 \pi}{180})\\\\v = 5 (\frac{4}{3})(\frac{\pi}{90})$

$v= \frac{2 \pi}{27} \ mi/s$

Thus,the velocity of the missile when the angle of elevation from the tracking station is 30 degrees = $\frac{2 \pi}{27} \ mi/s$