## Northern Component 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?

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 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.)
 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....

## Northern Component of velocity

ooo i have to do sin/cosine of the angle

 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

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

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

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

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