How Can Vector Addition Determine the Single Putt Needed in Golf?

[featherguy]

Homework Statement

A golfer is a putting on a green, and takes 3 strokes to "hole the ball" On the first putt the ball rolls 5.0m due east. The second putt travels 2.1 m at an angle of 20.0 degrees north of east. The third putt is 0.50m due north. What displacement (magnitude and direction relative to due east) would have been needed to "hole the ball" on the very first putt?

Homework Equations

R^2 = A^2 + B^2 - 2ab COS(angle)

The Attempt at a Solution

I drew them out. I have troubles finding the displacement with 3 directions given.

 

[LowlyPion]

Homework Statement

A golfer is a putting on a green, and takes 3 strokes to "hole the ball" On the first putt the ball rolls 5.0m due east. The second putt travels 2.1 m at an angle of 20.0 degrees north of east. The third putt is 0.50m due north. What displacement (magnitude and direction relative to due east) would have been needed to "hole the ball" on the very first putt?

Homework Equations

R^2 = A^2 + B^2 - 2ab COS(angle)

The Attempt at a Solution

I drew them out. I have troubles finding the displacement with 3 directions given.

Welcome to PF.

In problems like this I find it a little easier to separate them into x,y components and then sum the components.

So for your problem taking East as positive x ...

P1 = 5 x + 0y
P2 = 2.1*cos20 x + 2.1*sin20 y
P3 = 0 x + .5 y

Your resultant then is the sum of the 3 P vectors right? And your answers are precise. Easy peasy, nice and easy.

 

[nlightner]

I would suggest using the Pythagorean theorem to solve this problem. Since the displacement is a vector quantity, we can use the magnitude and direction of each putt to find the overall displacement.

First, we can calculate the magnitude of the first putt using the distance formula: d = √(x^2 + y^2), where x = 5.0m and y = 0m (since the putt is due east). This gives us a magnitude of 5.0m for the first putt.

Next, we can use the angle and magnitude of the second putt to find its x and y components. Using trigonometry, we can calculate that the x component is 2.1cos(20) = 1.994m and the y component is 2.1sin(20) = 0.723m.

For the third putt, we know that it is only 0.50m due north, so its x component is 0m and its y component is 0.50m.

Now, we can use vector addition to find the overall displacement. Adding the x components, we get 5.0m + 1.994m + 0m = 6.994m. Adding the y components, we get 0m + 0.723m + 0.50m = 1.223m.

Using the Pythagorean theorem, we can find the magnitude of the overall displacement: d = √(6.994m^2 + 1.223m^2) = 7.074m.

Finally, to find the direction of the overall displacement, we can use the inverse tangent function: θ = tan^-1(1.223m/6.994m) = 9.9 degrees north of east.

Therefore, the overall displacement needed for the ball to "hole" on the first putt would be 7.074m at an angle of 9.9 degrees north of east.