A golfer, putting on a green, requires three strokes to "hole the ball." During the first putt, the ball rolls 5.8 m due east. For the second putt, the ball travels 2.3 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?
ylkjyl
To calculate the displacement that would have been needed to "hole the ball" on the very first putt, we need to find the total horizontal and vertical displacements.
Let's break down the given information:
First putt:
- Horizontal displacement: 5.8 m due east
- Vertical displacement: 0
Second putt:
- Horizontal displacement: 2.3 m * cos(20°) = 2.3 m * 0.9397 = 2.161 m
- Vertical displacement: 2.3 m * sin(20°) = 2.3 m * 0.3420 = 0.7866 m (positive because it's going north)
Third putt:
- Horizontal displacement: 0
- Vertical displacement: 0.50 m (positive because it's going north)
Now, let's add up the horizontal and vertical displacements:
Horizontal displacement = 5.8 m + 2.161 m + 0 = 7.961 m
Vertical displacement = 0 + 0.7866 m + 0.50 m = 1.2866 m
To find the magnitude of the total displacement, we can use the Pythagorean theorem:
Magnitude of displacement = sqrt((horizontal displacement)^2 + (vertical displacement)^2)
= sqrt((7.961 m)^2 + (1.2866 m)^2)
= sqrt(63.2976 m^2 + 1.65523 m^2)
= sqrt(64.95283 m^2)
= 8.07 m (rounded to two decimal places)
To find the direction of the displacement relative to due east, we can use trigonometry:
Direction = tan^(-1)(vertical displacement / horizontal displacement)
= tan^(-1)(1.2866 m / 7.961 m)
= tan^(-1)(0.1614)
= 9.2° (rounded to one decimal place)
Therefore, the displacement that would have been needed to "hole the ball" on the very first putt is approximately 8.07 m in magnitude, at an angle of 9.2° north of east.