Comments on: determine the velcity (magnitude and dierction) of the wind with respect to ground? http://diy-solarpanels.net/air-x-wind/determine-the-velcity-magnitude-and-dierction-of-the-wind-with-respect-to-ground Build your own solar panels for less then $200! Tue, 07 Sep 2010 12:34:58 -0500 http://wordpress.org/?v=2.8.4 hourly 1 By: lonelyspirit http://diy-solarpanels.net/air-x-wind/determine-the-velcity-magnitude-and-dierction-of-the-wind-with-respect-to-ground/comment-page-1#comment-495 lonelyspirit Fri, 27 Nov 2009 22:17:59 +0000 http://diy-solarpanels.net/air-x-wind/determine-the-velcity-magnitude-and-dierction-of-the-wind-with-respect-to-ground#comment-495 The components of the wind which changed the direction of the plane to 45 degrees west of south is the resultant of the wind direction and the airplane heading. According to your data, I understand the solution as follows: The airplane had a y component = 180 degrees or heading south. There is no X component to the airplane during its flight before it was subject to the wind forces The Velocity of the airplane while the wind blowing for 900 seconds = 81000/900 = 90 m/second in the direction of 45 degrees west of south. The y component of that movement is V cos theta + 57.8 cos 45 = 0 The x component is v sin theta + 57.8 sine 45 = 90 m/s V cos theta = 41 V sin theta = 90 -40.5 = 49.5 Tan theta = 49.5/41 =1.22 . Theta =50.6 V cos 50.6 = 41, v = 41/0.63 =65 m/s at a direction 50.6 west of south.<br><b>References : </b><br> The components of the wind which changed the direction of the plane to 45 degrees west of south is the resultant of the wind direction and the airplane heading.
According to your data, I understand the solution as follows:

The airplane had a y component = 180 degrees or heading south. There is no X component to the airplane during its flight before it was subject to the wind forces

The Velocity of the airplane while the wind blowing for 900 seconds = 81000/900 = 90 m/second in the direction of 45 degrees west of south. The y component of that movement is V cos theta + 57.8 cos 45 = 0
The x component is v sin theta + 57.8 sine 45 = 90 m/s
V cos theta = 41
V sin theta = 90 -40.5 = 49.5
Tan theta = 49.5/41 =1.22 . Theta =50.6
V cos 50.6 = 41, v = 41/0.63 =65 m/s at a direction 50.6 west of south.
References :

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