Golf Displacement: 4.19m, 20° North of East

[Caitlin.Lolz.]

Homework Statement

A golfer, putting on a green, requires three strokes to "hole the ball." During the first putt, the ball rolls 4.0 m due east. For the second putt, the ball travels 3.0 m at an angle 20° north of east. The third putt is 0.50 m due north. What displacement (magnitude and direction relative to due east) would have been needed to "hole the ball" on the very first putt?
Magnitude, in meters, and Direction, degrees North of East.

Homework Equations

X = Vxt (I think)

The Attempt at a Solution

I tried to to draw a diagram of a vector but if I followed the directions exactly then the picture isn't a triangle. So, I attempted to find the X and Y components, but it doesn't seem right.
X) 3cos(20) + .50cos(90) + 4cos(70) = 4.19
I don't know what to do from finding the x and y components, or if I'm even finding the components correctly.

 

[LowlyPion]

Homework Statement

A golfer, putting on a green, requires three strokes to "hole the ball." During the first putt, the ball rolls 4.0 m due east. For the second putt, the ball travels 3.0 m at an angle 20° north of east. The third putt is 0.50 m due north. What displacement (magnitude and direction relative to due east) would have been needed to "hole the ball" on the very first putt?
Magnitude, in meters, and Direction, degrees North of East.

Homework Equations

X = Vxt (I think)

The Attempt at a Solution

I tried to to draw a diagram of a vector but if I followed the directions exactly then the picture isn't a triangle. So, I attempted to find the X and Y components, but it doesn't seem right.
X) 3cos(20) + .50cos(90) + 4cos(70) = 4.19
I don't know what to do from finding the x and y components, or if I'm even finding the components correctly.

If the golfer 3-putted it won't be a triangle, unless he was putting from the bottom of the hole.

You are on the right track ... sort of.
You need to separate each of the three vectors into x and y-axis components and then sum x and sum y separately.
That will be the resulting x and y component of the vector of the answer.
The magnitude will be given by the square root of the sum of the squares. And the direction by the angle that makes with a reference direction.

 

[Caitlin.Lolz.]

Ok - thank you. But, to find the x and y components don't I need to know the angles that go with each distance? How do I find the angle for 4 m East? or .50 m North? Thats mainly what I got stuck on. =].

 

[LowlyPion]

Ok - thank you. But, to find the x and y components don't I need to know the angles that go with each distance? How do I find the angle for 4 m East? or .50 m North? Thats mainly what I got stuck on. =].

Due east is a direction. If you choose your axis well you can eliminate terms.

In this regard choosing East as the positive x-axis should be a good choice. Then you know the x component is 4 and the y component is 0.

Once you define East then you should know where North is correct?

 

[Caitlin.Lolz.]

Thank you, that makes sense! And I got the problem right!

 

[LowlyPion]

Thank you, that makes sense! And I got the problem right!

Good job then. Keep up the good work.