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The 5.0-m-long rope in (Figure 1) 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. She runs as fast as she can, grabs the rope, and swings out over the ravine.
When she's directly over the far edge of the ravine, how much higher is she than when she started?
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?

Respuesta :

ajeigbeibraheem

She is 1.0 m higher than she is when she started. The speed at which the woman must be running when she grabs the rope is 4.427 m/s

Suppose we consider the Figure 1 to be a right-angle triangle;

where;

  • the hypotenuse is the rope = (5.0 m),
  • the opposite side is the ravine = (3.0 cm) and;
  • the adjacent is the slant imaginary distance = (x)

∴

Using Pythagoras rule;

(hyp)² = opp² + adj²

5.0² = 3.0² + x²

x² = 25 - 9

x² = 16

x = √16

x = 4 m.

When she is directly over the far edge of the ravine, the height at which she is far higher than where she started is:

= 5cm - 4cm

= 1 cm

The speed at which the woman must be running when she grabs the rope can be determined by taking both the potential energy and kinetic energy of the system, which can be computed as:

  • mgh = 1/2 mv²

Making the speed the subject of the formula, we have:

[tex]\mathbf{v = \sqrt{2gh}}[/tex]

[tex]\mathbf{v = \sqrt{2\times 9.8 \times 1.0}}[/tex]

v = 4.427 m/s

Learn more about kinetic energy here:

https://brainly.com/question/12669551?referrer=searchResults

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