physics
posted by bobjahng on .
In a World Cup soccer match, Juan is running due north toward the goal with a speed of 7.80 m/s relative to the ground. A teammate passes the ball to him. The ball has a speed of 13.6 m/s and is moving in a direction of 32.5 degrees east of north, relative to the ground.
a)What is the magnitude of the ball's velocity relative to Juan?
b)What is the direction of the ball's velocity relative to Juan?
How do you start this off? I found the y component and subtracted it to find the answer for the first part, but it says that my answer is wrong.

a) ball relativeground=Juansspeed+ball relative to Juan
13.6Cos32.5N + 13.6Sin32.5E =7.8N+ballrelative to Juan
ball relative to Juan= (13.6Cos32.57.8)N+13.6Sin32.5E
so figure out the N component, then magnitude= sqrt (Ncomp^2+Ecomp^2) 
i got the right answer for part a (8.18)
now how do i do part b? 
you have the vector:
Ncomp, E component
relative to Juan, going N, so it is E of him by theta, where tantheta=Ecomp/Ncomp 
i cant get part b!!!!!!! im completely lost >.<

Hi Chris

On that Kobe System.
Or Linsanity 
Make sure you're putting the parenthesis in the correct place. Such as for part A, the formulas should be:
a) (ball relative ground) = (Juan's speed)+ (ball relative to Juan)
(13.6Cos(32.5))N + (13.6Sin(32.5))E =(7.8)N+(ball relative to Juan)
ball relative to Juan= (13.6(Cos(32.5))7.8)N+(13.6Sin(32.5))E
so figure out the N component, then magnitude= sqrt (Ncomp^2+Ecomp^2)