In Figure 4-55, a radar station detects an airplane approachingdirectly from the

Question

In Figure 4-55, a radar station detects an airplane approachingdirectly from the east. At first observation, the airplane is atdistance d1 = 350 m from the station and atangle 1 = 38° above the horizon. The airplaneis tracked through an angular change = 117° in thevertical east–west plane; its distance is thend2 = 800 m. Find the (a)magnitude and (b) direction of the airplane'sdisplacement during this period. Give the direction as an anglerelative to due west, with a positive angle being above the horizonand a negative angle being below the horizon

Explanation / Answer

I had the same problem but with differentnumbers. Just plug in your numberss. A radar station detects an airplaneapproaching directly from the east. At first observation, the rangeto the plane is d1 = 360 m at 40° above the horizon. Theairplane is tracked for another 123° in the vertical east-westplane, the range at final contact being d2 = 790 m. See Fig. 5-35.Find the displacement of the airplane during the period ofobservation.
Magnitude


Point 1 is at a point on a triangle aboveanother point on the ground below it and East of the radarstation.

The elevation of point 1 is 40 deg and range is along the directline from radar to plane.

This triangle yields the plane's height as
y1 = 360 * sin(40deg)
and its ground distance from the radar station as
x1 = 360 * cos(40deg)

The elevation of point 2 is 180 - 123 - 40 deg or 17 deg

Point 2 is at a point on a triangle above another point on theground below it and West of the radar station.
This triangle yields the plane's height as
y2 = 790 * sin(17deg)
and its ground distance from the radar station as
x2 = -790 * cos(17deg) the minus is important!

The distance between points 1 & 2 is:
d = sqrt{ (x1 - x2)^2 + (y1 + y2)^2}
NOTE the negative for x2!

Now take out your calculator and complete the puzzle. Point 1 is at a point on a triangle aboveanother point on the ground below it and East of the radarstation.

The elevation of point 1 is 40 deg and range is along the directline from radar to plane.

This triangle yields the plane's height as
y1 = 360 * sin(40deg)
and its ground distance from the radar station as
x1 = 360 * cos(40deg)

The elevation of point 2 is 180 - 123 - 40 deg or 17 deg

Point 2 is at a point on a triangle above another point on theground below it and West of the radar station.
This triangle yields the plane's height as
y2 = 790 * sin(17deg)
and its ground distance from the radar station as
x2 = -790 * cos(17deg) the minus is important!

The distance between points 1 & 2 is:
d = sqrt{ (x1 - x2)^2 + (y1 + y2)^2}
NOTE the negative for x2!

Now take out your calculator and complete the puzzle. Point 1 is at a point on a triangle aboveanother point on the ground below it and East of the radarstation.

The elevation of point 1 is 40 deg and range is along the directline from radar to plane.

This triangle yields the plane's height as
y1 = 360 * sin(40deg)
and its ground distance from the radar station as
x1 = 360 * cos(40deg)

The elevation of point 2 is 180 - 123 - 40 deg or 17 deg

Point 2 is at a point on a triangle above another point on theground below it and West of the radar station.
This triangle yields the plane's height as
y2 = 790 * sin(17deg)
and its ground distance from the radar station as
x2 = -790 * cos(17deg) the minus is important!

The distance between points 1 & 2 is:
d = sqrt{ (x1 - x2)^2 + (y1 + y2)^2}
NOTE the negative for x2!

Now take out your calculator and complete the puzzle.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.