Question on displacement vectors

Question

A golfer, putting on a green, requires three strokes to "hole the ball." During the first putt, the ball rolls 5.0 m due east. For the second putt, the ball travels 2.1 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?

Answer 1

The resultant vector can be found algebraically via the following 3 steps

1. Find the x and y components of each vector.

2. Find the x and y components of the resulting vector by adding the corresponding values in step 1

3. Calculate the magnitude and direction from the x and y components of the resulting vector found in step 2.

 

For this problem

Step 1.

V₁ = 5i + 0j = <5, 0>

V₂ = 2.1 [cos(20°) i + sin(20°) j] = <2.1 cos(20°), 2.1 sin(20°)>

     = 1.97 i + 0.72 j = < 1.97, 0.72>

V₃ = 0i + 0.5j = <0, 0.5>

 

Step 2.

R = (5 + 1.97 + 0)i + (0 + 0.72 + 0.5)j = <5 + 1.97 + 0, 0 + 0.72 + 0.5>

   = 6.97i + 1.22j = <6.97, 1.22>

 

Step 3.

|R| = SQRT(6.97² + 1.22²) = 7.08 m

θ    = tan^-1(1.22/6.97) = 9.93°

 

So to "hole the ball" on the very first putt, the ball must travel 7.08 m at an angle 9.93° north of east.

 

Note:

Not sure which vector notation that you learn in your class, so I use both vector notations.

Answer 2

20 degrees north of east is an unusual description but it is just a conventional angle of 20 degrees.  Convert all vectors to components

 

putt 1   < 5.0, 0.0>

putt 2  < 2.1 cos 20, 2.1 sin 20>

putt 3  < 0, 0.5>

 

Add components and find the resultant vector

<6.973, 1.218>    Using trigonometry we can find that the resultant vector in polar coordinates is 7.087 at 9.90 degrees north of east.