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DochEvi [55]
3 years ago
15

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 el

evation 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.
Physics
1 answer:
Tanzania [10]3 years ago
7 0

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

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