A novice golfer on the green takes three
strokes to sink the ball. The successive dis-
placements are 7
.
7 m to the north, 1
.
9 m 45
◦
north of east, and 8
.
7 m 76
◦
west of south.
Starting at the same initial point, an expert
(lucky) golfer could make the hole in a single
displacement.
What is the magnitude of this single dis-
placement?
Disp. = 7m[90o] + 9m[E45oN] + 7m[S76oW].
X = 9*Cos45 - 7*Cos76 = 1.84 m.
Y = 7 + 9*sin45 - 7*sin76 = 6.57 m.
Q1.
Tan A = Y/X = 6.57/1.84 = 3.57168.
A = 74.4o N. of E. = Direction.
Disp. = Y/sin A = 6.57/sin74.4 = 6.82 m
Disp. = 7m[90o] + 9m[E45oN] + 7m[W14oS]
X = 9*Cos45 + (-7*Cos14) = -0.428 m.
Y = 7 + 9*sin45 + (-7*sin14) = 11.7 m.
Q2.
Tan A = Y/X = 11.7/-0.428 = -27.26754.
A = -87.9o = 87.9o N. of W. = Direction.
Disp. = Y/sin A = 11.7/sin87.9 = 11.71 m
To find the magnitude of the single displacement made by the expert golfer, we can use the concept of vector addition. We need to add the three given displacements together to get the single displacement.
The given displacements are as follows:
1. 7.7 m to the north
2. 1.9 m at 45° north of east
3. 8.7 m at 76° west of south
Let's break down the second displacement into its horizontal and vertical components.
Using trigonometry, we can find that:
Horizontal component = 1.9 m * cos(45°)
Vertical component = 1.9 m * sin(45°)
Now we can add the displacements. To do this, we treat the displacements as vectors and add them together component-wise.
Let's denote the north direction as positive y and the east direction as positive x.
Horizontal component = 1.9 m * cos(45°) = 1.34 m (east)
Vertical component = 1.9 m * sin(45°) = 1.34 m (north)
Adding the components together:
North: 7.7 m + 1.34 m = 9.04 m
East: 1.34 m + (-8.7 m) = -7.36 m (west)
Finally, we can find the magnitude of the single displacement using the Pythagorean theorem:
Magnitude = sqrt((9.04 m)^2 + (-7.36 m)^2)
By calculating this expression, we can find the magnitude of the single displacement made by the expert golfer.