+27 
+291 correct me if im wrong, always convert it into meters/seconds when doing physics, just divde it by 3.6
g=ground velocity
a=airplane velocity
w= wind velocity
the general equation would be V(g)=V(a)+V(w)
since they're on angles you would have to split them up into 2 components x and y so;
x; V(g)= 138.8cos(180-45)+33.3cos(180+25)
y; V(g)= 138.8sin(180-45)+33.3sin(180+25)
solve them when you get both components you will have to find the vector between them so;
|V(g)|=sqr(t(x^2+y^2))
and you should get your vector
hope it helped :)
+291 correct me if im wrong, always convert it into meters/seconds when doing physics, just divde it by 3.6
g=ground velocity
a=airplane velocity
w= wind velocity
the general equation would be V(g)=V(a)+V(w)
since they're on angles you would have to split them up into 2 components x and y so;
x; V(g)= 138.8cos(180-45)+33.3cos(180+25)
y; V(g)= 138.8sin(180-45)+33.3sin(180+25)
solve them when you get both components you will have to find the vector between them so;
|V(g)|=sqr(t(x^2+y^2))
and you should get your vector
hope it helped :)
I would leave it in km/hr since the plane's speed and wind are in the same units.
(one thing to be careful with here is the wind direction.....is it BLOWING in the direction of 250 degrees or is it COMING FROM 250 degrees.....as a pilot the latter would be more likely correct, but the way the question is worded it sounds like the former....the resultant ground speed will be the same BUT the resultant angle of the vector will be DIFFERENT) I will proceed with the wind BLOWING in the direction of 250 degrees:
Break both the plane and wind speeds into vertical and horizontal components
and sum them up.
Planex = 500 cos 135 = - 353.5533
Planey = 500 sine135 = 353.5533
Windx = 120 cos 250 = - 41.042
Windy = 120 sine250 = -112.764
Resultant x = -353.5533 - 41.024 = -394.577
Resultant y = 353.5533 - 112.764 = 240.789
Use pythagorean theorem x^2 + y^2 = resultant^2 Resultant = 462.245 KM/HR
With the x and y resultants you can figure out the new angle too (if the pilot does not correct by turning into the wind and 'crabbing' a bit) .....do you know how? (HINT: Tangent)
~jc
(hope I did all of the calculator math correctly!)
+129847 Ground speed can be determined by the vector sum of the aircraft's true airspeed and the current wind speed and direction
The plane is flying NW = 135°
The wind is blowing W25S = West and 25° South = 205°
The x vector component of the resultant is
500cos(135) + 120cos(205) = about -462.31
The y component of the resultant is
500sin135) + 120sin(205) = about 302.84
So the magnitude of the resultant vector [ the groundspeed ) = sqrt [ (-462.31)^2 + (302.84)^2] = about 552.67km/hr
And the angle of the resultant = tan-1(302.84/-462.31) = -33.23° + 180° = about 146.77° = about N56.77W
