# Physics

posted by on .

I gues I'm looking at this the wrong way i do not understand the work you do to get to the answer

Question:

Find the air velocity of a plane that must have a realtive ground velocity of 250.0 h^-1 km norht if it encounters a wind pushing it toward the northeast at 75 h^-1 km

The Work:

I do not follow it one bit...

Va = Vg - Vw
= 250 h^-1 km north - 75.0 h^-1 km northeast
= 250.0 h^-1 km north + 75.0 h^-1 km southwest
Va^2 = Vg^2 + Vw^2 - 2VgVw cos A
= (250.0 h^-1 km)^2 + (75.0 h^-1 km)^2 - 2(250.0 h^-1 km)(75.0 h^-1 km)(cos 45.0)
= 62 500 h^-2 km^2 + 5 625 h^-2 km^2 + 26 516 h^-2 km62
= 41 609 h^-2 km^2
Va = 204 h^-1 km
Vw/sin W = Va sin A
sin W = Vw sin A/Va
= (75 h^-1 km)(sin 45.0)/(204 h^-1 km)
= (75 h^-1 km)(.707)/(204 h^-1 km)
= 0.260
B = 15.1
Answer = 204 h^-1 km, 15.1 degrees west of north

ok i was lost on line 2 and three were the added and subtracted 75 with two different directions why are there two

line 3 were did this come from

- 2VgVw cos A

thanks!

• Physics - ,

Pythagoras theorem applies to the sides adjacent to the right angle:
a²+b²=c²
The extension when the angle is not a right angle is called the cosine rule, which states that:
a²+b²-2ab cos(C) = c²
which is what line three is doing, the angle between NE and N being 45°.
The resultant (air velocity) was calculated at 204 km/hr.

The line
Vw/sin W = Va sin A
Vw/sin W = Va/sin A
which is the application of the sine rule to find the angle of a triangle in which one side and the opposite angle is known, as well as the side opposite to the angle sought.

I have not checked the numerical calculations. You can do the checking as an exercise.