A golfer, putting on a green, requires three strokes to \"hole the ball.\" Durin
ID: 1682843 • Letter: A
Question
A golfer, putting on a green, requires three strokes to "hole the ball." During the first putt, the ball rolls 4.07 m due east. For the second putt, the ball travels 2.04 m at an angle of 35.0 ° north of east. The third putt is 0.382 m due north. What is the magnitude of the displacement that would have been needed to "hole the ball" on the very first putt?A golfer, putting on a green, requires three strokes to "hole the ball." During the first putt, the ball rolls 4.07 m due east. For the second putt, the ball travels 2.04 m at an angle of 35.0 ° north of east. The third putt is 0.382 m due north. What is the magnitude of the displacement that would have been needed to "hole the ball" on the very first putt?
Explanation / Answer
A golfer, putting on a green, requires three strokes to "hole the ball." During the first putt, the ball rolls 4.07 m due east. For the second putt, the ball travels 2.04 m at an angle of 35.0 ° north of east. The third putt is 0.382 m due north. What is the magnitude of the displacement that would have been needed to "hole the ball" on the very first putt? First x1 = 4.07 m , y1 = 0 m x2 = 2.04 cos35 , y2 = 2.04 sin35 x3 = 0 m , y3 = 0.382 m So we have x = x1 + x2 + x3 = 4.07 + 2.04 cos35 + 0 = 5.741 m y = y1 + y2 + y3 = 0 + 2.04 sin35 + 0.382 = 1.552 m Therefore magnitude R = [x^2 + y^2]^1/2 = 5.947 m direction theta = tan-(y/x) = 15.12 deg