Physics ch 6 ex10???

Question

The 5.0 m-long rope in the figure hangs vertically from a tree right at the edge of a ravine. A woman wants to use the rope to swing to the other side of the ravine, which is 3m wide. She runs as fast as she can, grabs the rope, and swings out over the ravine.

1)When she's directly over the far edge of the ravine, how much higher is she than when she started?

2)Given your answer to Part A, how fast must she be running when she grabs the rope in order to swing all the way across the ravine?

Answer 1

Because you didn't provide the figure, I am going to assume that the woman grabs the end of the rope.

1) On the figure, draw the rope when the woman is over the far side of the ravine. Draw a horizontal line segment from the end of the rope in this position, back to the vertical line representing the rope before the woman swings on it, meeting that line at point P. This horizontal line segment has length 3 m, and the rope still has length 5 m, so the distance from P to the place where the rope is tied to the tree is 4 m (You will have drawn a 3-4-5 right triangle.) Therefore, she is 1 meter higher than when she started.

2) The minimum speed she must be running when she grabs the rope is the speed such that her kinetic energy, converted to gravitational potential energy, is just enough to raise her 1 meter. Therefore

½mv² = mgh

v = √(2gh)

I'll let you do the arithmetic.