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

the velocity of the missile when the angle of elevation from the tracking station is 30 degrees =
`(2 \pi)/(27) \ mi/s`

Step-by-step 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 = (h)/(5)\\\\h = 5 tan \ \theta`

Differentiating the above equation ; we have

`(dh)/(dt) = 5 \ sec^2 \ \theta \ (d \theta)/(dt)`

Replacing
`v \ with \ (dh)/(dt)` ;
`\theta \ with \ 30^0` and
`(d\theta)/(dt) \ with \ \ (2 \pi)/(180)`; we have :

`v = 5 \ sec^2 \ (30^0) \ ((2 \pi)/(180))\\\\v = 5 ((4)/(3))((\pi)/(90))`

`v= (2 \pi)/(27) \ mi/s`

Thus,the velocity of the missile when the angle of elevation from the tracking station is 30 degrees =
`(2 \pi)/(27) \ mi/s`