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A golfer takes three strokes to putt a golf ball into a hole. On the first stroke, the ball moves 4.8 m due east. On the second, it moves 2.7 m at an angle 20° north of east. On the third, it moves 0.50 m due north. If the golfer had instead hit the ball directly into the hole on the first stroke, what would have been the magnitude (in m) and direction of the ball's displacement? Give the direction as a positive angle measured counterclockwise from due east.

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fmorelos

Answer:

7.47 m, 10.98° north of east.

Explanation:

You need to add the vectors and find the resultant. You can do this using the component method, where you add together the x-component and y-component of each vector:

First stroke:

[tex]D_1_x = 4.8m\\D_1_y = 0m[/tex]

Second stroke:

[tex]D_2_x = 2.7m*cos(20)\\D_2_y = 2.7m*sin(20)[/tex]

Third stroke:

[tex]D_3_x = 0m\\D_3_y = 0.5m[/tex]

Summation:

[tex]D_x = 4.8m + 2.7m*cos(20) + 0m = 7.34m\\D_y = 0m + 2.7m*sin(20) + 0.5m = 1.42m[/tex]

The magnitude is found using Pythagorean theorem, the direction is found using trigonometry:

[tex]D = \sqrt{D_x^2 + D_y^2} = 7.47 m\\\alpha = arctan(\frac{D_y}{D_x}) = 10.98[/tex]° north of east

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