Question: Explain how you would use the result of finding the ground speed to find the course of the airplane.
Answer: 181.2
Explanation: vertical speed of 20.78 knots
new course is (-148 , 20.78) = inverse tan ( 20.78 / -148) = 172° or 8° N of East at 149.45 knots
New speed (- 148)² + ( 20.28)² = 149.45 knots round and you get 181.2
question answered by
(jacemorris04)
The ground speed of the airplane is 181.18 knots (approx).
The formula of ground speed of a airplane is given below,
[tex]v_{g} =\sqrt{}[/tex][tex](v_{a} ^{2}+v_{w} ^{2}-2v_{a} v_{w}cos\alpha)[/tex]
where, [tex]v_{g}[/tex] = Ground speed of airplane
[tex]v_{a}[/tex] = Speed of the airplane relative to the air
[tex]v_{w}[/tex] = Wind speed
[tex]\alpha[/tex] = Internal angle
Given, [tex]v_{a}[/tex] = 160 knots, [tex]v_{w}[/tex] = 24 knots
Here, the angles increase clockwise and east is 90° heading.
The airplane is flying a heading of 90° at 160 knots speed & wind is blowing from 240° at 24 knots speed.
So, the internal angle[tex](\alpha)[/tex] = 240°-90° = 150°
∴ Ground speed, [tex]v_{g} =\sqrt{}[/tex][tex](v_{a} ^{2}+v_{w} ^{2}-2v_{a} v_{w}cos\alpha)[/tex]
⇒ [tex]v_{g}[/tex] [tex]=\sqrt{}[/tex][tex](160^{2}+24^{2}-2*160*24*cos(150))[/tex]
⇒ [tex]v_{g}[/tex] = 181.18 knots (approx)
Hence the ground speed of airplane is 181,18 knots (approx).
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