Solving Mechanics Problem: Cessna at Taupo Airport with Easterly Wind of 38 km/h

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A Cessna aircraft departs from Taupo Airport with an airspeed of 137 km/h at an angle of 26° East of North, facing an Easterly wind of 38 km/h. To solve the problem, a wind vector triangle is utilized, requiring geometric representation on a coordinate system. After calculating the resultant vector, the distance from Taupo after 50 minutes of flight is determined. The original poster successfully arrives at the correct answer with assistance from other forum members. The discussion emphasizes the importance of understanding vector components in aviation scenarios.
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On another day the Cessna leaves Taupo airport when there is an Easterly wind (i.e a wind blowing from direction East) of speed 38 km/h. If the plane starts out directly above Taupo with an airspeed v=137 km/h, pointing in a direction θ = 26 ° East of North, how far is it from Taupo after 50 min of flying at constant altitude?

i tried to do this problem but i cannot get a correct answer.. i tried all different ways but still i can't get it..
can u help please.
 
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Your question refers to what pilots call a 'wind vector triangle'. It belongs in the homework section, though. Welcome to PF.
 
You need to do some geometry. Draw a set axes, and mark them N, S, E and W. Now draw a line ( starting at the origin) at 28deg E of N with length 137 ( any units you like).
From the tip of this line draw another going E with length 38 ( same units).
Now connect the tip of this line with the origin ( Taupo).
The answer is the length of this line times 50 Km.
 
yup got it.

i got the answer

thanks a lot for your help.

thank u
 

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