Answer:
the velocity of the missile when the angle of elevation from the tracking station is 30 degrees = [tex]\frac{2 \pi}{27} \ mi/s[/tex]
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 ;
[tex]tan \ \theta = \frac{h}{5}\\\\h = 5 tan \ \theta[/tex]
Differentiating the above equation ; we have
[tex]\frac{dh}{dt} = 5 \ sec^2 \ \theta \ \frac{d \theta}{dt}[/tex]
Replacing [tex]v \ with \ \frac{dh}{dt}[/tex] ; [tex]\theta \ with \ 30^0[/tex] and [tex]\frac{d\theta}{dt} \ with \ \ \frac{2 \pi}{180}[/tex]; we have :
[tex]v = 5 \ sec^2 \ (30^0) \ (\frac{2 \pi}{180})\\\\v = 5 (\frac{4}{3})(\frac{\pi}{90})[/tex]
[tex]v= \frac{2 \pi}{27} \ mi/s[/tex]
Thus,the velocity of the missile when the angle of elevation from the tracking station is 30 degrees = [tex]\frac{2 \pi}{27} \ mi/s[/tex]
