### Physics Problem: Swinging on a Vine**Problem Statement:**Jane, looking for Tarzan, is running at a speed of 6.40 m/s and grabs a vine hanging vertically from a tall tree in the jungle. How high can she swing upward? \((g = 9.8 \, m/s^2)\)[Input Box]__________ meters**Detailed Explanation:**1. **Given Data:** - Initial speed (v_i): 6.40 m/s - Acceleration due to gravity (g): 9.8 m/s²2. **Concept Applied:** - The problem involves the conversion of kinetic energy to potential energy.3. **Formulas:** - Kinetic Energy (K.E): \( \frac{1}{2}mv_i^2 \) - Potential Energy (P.E): \( mgh \) - At the highest point, all kinetic energy will convert into potential energy: \[ \frac{1}{2}mv_i^2 = mgh \]4. **Solving for Height (h):** \[ h = \frac{v_i^2}{2g} \] Substituting the given values: \[ h = \frac{(6.40 \, m/s)^2}{2 \times 9.8 \, m/s^2} \] \[ h = \frac{40.96}{19.6} \] \[ h \approx 2.09 \, meters \]So, the height \( h \) that Jane can swing upward is approximately 2.09 meters.**Reminder:** This calculation assumes no energy loss due to air resistance or friction.

initial speed = V0 = 6.4 m/sheight = h = ?

Answered: Jane, looking for Tarzan, is running at…

Jane, looking for Tarzan, is running at a speed (6.40 m/s) and grabs a ne hanging vertically from a tall tree in the jungle. How high can she ving upward?(g = 9.8 m/s²)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

### Physics Problem: Swinging on a Vine

**Problem Statement:**

Jane, looking for Tarzan, is running at a speed of 6.40 m/s and grabs a vine hanging vertically from a tall tree in the jungle. How high can she swing upward? \((g = 9.8 \, m/s^2)\)

[Input Box]
__________ meters

**Detailed Explanation:**

1. **Given Data:**
- Initial speed (v_i): 6.40 m/s
- Acceleration due to gravity (g): 9.8 m/s²

2. **Concept Applied:**
- The problem involves the conversion of kinetic energy to potential energy.

3. **Formulas:**
- Kinetic Energy (K.E): \( \frac{1}{2}mv_i^2 \)
- Potential Energy (P.E): \( mgh \)
- At the highest point, all kinetic energy will convert into potential energy:
\[ \frac{1}{2}mv_i^2 = mgh \]

4. **Solving for Height (h):**
\[ h = \frac{v_i^2}{2g} \]

Substituting the given values:
\[ h = \frac{(6.40 \, m/s)^2}{2 \times 9.8 \, m/s^2} \]
\[ h = \frac{40.96}{19.6} \]
\[ h \approx 2.09 \, meters \]

So, the height \( h \) that Jane can swing upward is approximately 2.09 meters.

**Reminder:** This calculation assumes no energy loss due to air resistance or friction.

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