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 of 20.0 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

  First break each vector into the x-components ( ĭ ) and y-components ( ĵ ):
putt 1 = p1
putt 2 = p2
putt 3 = p3

p1 = 5.0ĭ + 0ĵ (meaning it moved 5 in the x-direction and 0 in the y-direction)
p2 = [2.1*sin(20)] ĭ + [2.1*cos(20)] ĵ (you have to draw the triangle to see it more clearly
p3 = 0ĭ + 0.5ĵ

the total putt, P is:
P = p1 + p2 + p3
P = (5.0ĭ + 0ĵ) + [2.1*cos(20)] ĭ + [2.1*sin(20)] ĵ + (0ĭ + 0.5ĵ)

now factor out the ĭ and ĵ from each putt.
P = [5.0 + 2.1*cos(20)] ĭ + [2.1*sin(20) + 0.5] ĵ

solve inside the brackets and get:
P = 6.97ĭ + 1.22ĵ

This is still broken down into x- and y- components. Simply use the pythagorean theorem to find the total vector.

P = √(6.97² + 1.22²) = 7.08m

To find the direction, do:
tanˉ¹(1.22 / 6.97) = 10° north of south

FINAL ANSWER: 7.08m, 10° north of south