Asked by JHM
A plane head north with an airspeed of 420 km/h. However, relative to the ground, it travels in a direction 7.0 west of north. if the wind's direction is towards the southwest (45.0 south of west), what is the wind speed?
Answers
Answered by
Henry
All angles are CCW from +x-axis.
Vp + Vw = 420km/h[97o].
420[90o] + Vw[225o] = 420[97o]
420i + Vw*Cos225+Vw*sin225 = 420*Cos97 +420*sin97 = 420i - 0.707Vw - i0.707Vw = -51.2 + 417i = -51.2 -3i.
0.707Vw (-1-1i) = -51.2 - 3i.
0.707Vw(1.41[225o]) = 51.3[183.4o].
Vw[225o] = 51.3[183.4o]
Vw = 51.3km/h[-41.6]
Speed of wind = 51.3km/h.
Vp + Vw = 420km/h[97o].
420[90o] + Vw[225o] = 420[97o]
420i + Vw*Cos225+Vw*sin225 = 420*Cos97 +420*sin97 = 420i - 0.707Vw - i0.707Vw = -51.2 + 417i = -51.2 -3i.
0.707Vw (-1-1i) = -51.2 - 3i.
0.707Vw(1.41[225o]) = 51.3[183.4o].
Vw[225o] = 51.3[183.4o]
Vw = 51.3km/h[-41.6]
Speed of wind = 51.3km/h.
Answered by
anya
vector diagram would be the 420 heading north + unknown head NW = unknown heading 7 degrees W of N
7 degrees in bottom corner 45 + 90= 135 above that and 180-7-135 = 38 in the top left corner
using sine law Vw/sin(7)= 420/sin(38) where Vw = velocity of wind = 83km/hr
7 degrees in bottom corner 45 + 90= 135 above that and 180-7-135 = 38 in the top left corner
using sine law Vw/sin(7)= 420/sin(38) where Vw = velocity of wind = 83km/hr
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