# Northern Component of velocity

1. Sep 12, 2007

### anglum

Components of velocity

1. The problem statement, all variables and given/known data

An airplane travels at 146 km=h toward the
northeast.
What is the northern component of its ve-
locity? Answer in units of km=h.

2. Relevant equations

asquared + bsquared = csquared

3. The attempt at a solution

however i am not sure if u use the pythagorean formula to solve this?

Last edited: Sep 12, 2007
2. Sep 12, 2007

### drpizza

Yes, you're on the right track. (Assuming by "North East" they mean that 45 degree angle between north and east. That means NorthEast would be the hypotenuse/resultant of a two vectors: 1 north and 1 east of equal magnitudes.)

3. Sep 12, 2007

### Staff: Mentor

First, the plane speed is 146 km/h (not km=h).

Next, you find the components in the north and east directions by multiplying the hypoteneuse by the sine or cosine of the appropriate angles....

4. Sep 12, 2007

### anglum

ooo i have to do sin/cosine of the angle

5. Sep 12, 2007

### anglum

so to find the horizontal velocity it would be the cos of 45 = vx/ 146? which equals 76.699 km/h

and to get the vertical it would be sin of 45 = vy/146? which equals 124.231 km/h

Last edited: Sep 12, 2007
6. Sep 12, 2007

### Staff: Mentor

Yes. That's an unusual way to write it, however. More like this (I'll use latex):

$$v_y = 146 km/h * sin(45)$$

7. Sep 12, 2007

### Staff: Mentor

Well, except for the math you did. The sin and cos of 45 degrees should be the same....

8. Sep 12, 2007

### anglum

ok thanks guys i got it i was doing some bad math... thank you

Last edited: Sep 12, 2007