Northern Component of velocity

anglum

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?

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.)

berkeman

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....

anglum

ooo i have to do sin/cosine of the angle

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

berkeman

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

[tex]v_y = 146 km/h * sin(45)[/tex]

berkeman

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

anglum

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