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A golf ball lies 2.00 m directly south of the hole on a levelgreen. On the first putt, the ball travels 3.00 m along astraight-line path at an angle o

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asked 2021-05-05
A golf ball lies 2.00 m directly south of the hole on a levelgreen. On the first putt, the ball travels 3.00 m along astraight-line path at an angle of 5 degrees east of north; on thesecond putt, it travels a straight-line distance of 1.20 m at anangle of 6 degrees south of west.
What would be the displacement ofa third putt that would put the ball in the hole?
The answer in the back of the book is 1.3 m at 43 degreessouth of east. How do I get that?

Answers (1)

2021-05-07

Add the vector components of the putts:
On the firstputt \(x_1\)= 3 m * \(\displaystyle{\sin{{\left({5}\right)}}}\)= .26m and \(y_1\) = 3m * \(\displaystyle{\cos{{\left({5}\right)}}}\) = 2.989 m
on the second putt \(x_2\) = 1.2 m *\(\displaystyle{\cos{{\left({186}\right)}}}\) = -1.19 m and \(y_2\) = 1.2 m * \(\displaystyle{\sin{{\left({186}\right)}}}\) = -1.25 m
The resulting components are X = -.93 m and Y = 2.864 m
Since the hole originally was at 2 m N the Y component of theball from the hole is .864 m
So the resultant distance to the hole - \(\displaystyle{\left(-{.93}^{{{2}}}+{.864}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}={1.27}{m}\)
Also, the ball is .93 m W of the hole and .864 m N of the holeand tan = .864 / .93
and = 42.9 deg

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