An airplane takes off at a ground velocity of 200km/hr toward the east, climbing at an angle of 14 degrees. A 20km/hr wind is blowing from the north. Determine the resultant air and ground velocities and their magnitudes.

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- Jun 1st 2009, 02:39 PMdalbir4444Finding resultants in 3-dimensions
An airplane takes off at a ground velocity of 200km/hr toward the east, climbing at an angle of 14 degrees. A 20km/hr wind is blowing from the north. Determine the resultant air and ground velocities and their magnitudes.

- Jun 1st 2009, 10:02 PMearboth
Use a coordinate system with the x-axis pointing East, the y-axis pointing North and the z-axis pointing up into the sky.

Then the North wind can be described by:

$\displaystyle \vec w=(0, -20, 0)$

and the starting airplane as

$\displaystyle \vec p=(200\cdot \cos(14^\circ)\ ,\ 0\ ,\ 200\cdot \sin(14^\circ))$

The resulting vector which described the way of the airplane is:

$\displaystyle \vec r = \vec w + \vec p = (200\cdot \cos(14^\circ)\ ,\ -20\ ,\ 200\cdot \sin(14^\circ))$

and it's magnitude is:

$\displaystyle |\vec r|=\sqrt{(200\cdot \cos(14^\circ))^2+(-20)^2+ (200\cdot \sin(14^\circ))^2}=\sqrt{40400} \approx 201\ \frac{km}h$

The course over ground is described by:

$\displaystyle \vec g = (0,-20)+(200\cdot \cos(14^\circ)\ ,\ 0)=(200\cdot \cos(14^\circ)\ ,\ -20)$ which yields

$\displaystyle |\vec g|=\sqrt{(200\cdot \cos(14^\circ))^2+(-20)^2}\approx 195.1\ \frac{km}h$ - Jun 2nd 2009, 03:08 AMdalbir4444
The back of my book says that the air velocity is [200, 20, 200 tan14] which gives a magnitude of 207.1 km/h.

Also, the ground velocity is given as[200,20,0] which gives a magnitude of 201.0 km/h.

I couldn't make sense of this answer. Should I assume that the book is wrong. - Jun 2nd 2009, 05:09 AMearboth
No, it's me who made the mistake. I overlooked that the ground speed of the airplane was given (I assumed that the speed through the air was 200 km/h).

So in short:

$\displaystyle \vec w=(0, -20, 0)$

$\displaystyle \vec p=(200\ ,\ 0\ ,\ 200\cdot \tan(14^\circ))$

$\displaystyle \vec r = \vec w + \vec p = (200\ ,\ -20\ ,\ 200\cdot \tan(14^\circ))$

$\displaystyle

|\vec r|=\sqrt{(200)^2+(-20)^2+ (200\cdot \tan(14^\circ))^2}=\sqrt{42886.578} \approx 207.1\ \frac{km}h$

And the speed over ground is then:

$\displaystyle

|\vec g|=\sqrt{(200)^2+(-20)^2}\approx 201\ \frac{km}h

$

Again: I'm sorry for the confusion!