14. A 75 kg pole vaulter running at 20 m/s vaults over the bar. If the vaulter's horizontal component of velocity over the bar is 15 m/s and air resistance is disregarded, how high was the jump?
14. A 75 kg pole vaulter running at 20 m/s vaults over the bar. If the vaulter's horizontal component of velocity over the bar is 15 m/s and air resistance is disregarded, how high was the jump?
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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How high was the jump?
![**Problem 14: Pole Vaulting Physics**
A 75 kg pole vaulter running at 20 m/s vaults over the bar. If the vaulter's horizontal component of velocity over the bar is 15 m/s and air resistance is disregarded, how high was the jump?
**Explanation:**
To solve this problem, we need to understand the conservation of energy and components of velocity involved:
- **Initial Kinetic Energy (KE):**
\[
KE_{\text{initial}} = \frac{1}{2} m v_{\text{initial}}^2
\]
where \( m = 75 \, \text{kg} \) and the initial running speed \( v_{\text{initial}} = 20 \, \text{m/s} \).
- **Final Kinetic Energy (Horizontal Component):**
\[
KE_{\text{horizontal}} = \frac{1}{2} m v_{\text{horizontal}}^2
\]
where \( v_{\text{horizontal}} = 15 \, \text{m/s} \).
- **Potential Energy (PE) at the top of the jump:**
\[
PE = mgh
\]
where \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity), and \( h \) is the height to be determined.
**Conservation of Energy:**
The initial kinetic energy is converted into potential energy at the top of the vault and the kinetic energy of the horizontal motion:
\[
\frac{1}{2} m v_{\text{initial}}^2 = \frac{1}{2} m v_{\text{horizontal}}^2 + mgh
\]
**Solving for \( h \):**
Rearrange the equation to find the height \( h \):
\[
h = \frac{\frac{1}{2} m v_{\text{initial}}^2 - \frac{1}{2} m v_{\text{horizontal}}^2}{mg}
\]
Simplify:
\[
h = \frac{\frac{1}{2} (v_{\text{initial}}^2 - v_{\text{horizontal}}^2)}{g}
\]
Substitute the known values:
\[
h = \frac{\frac{1}{2} (20^2 - 15^](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F51017fe1-0672-4cad-89e4-c6f0a0bf2651%2Fb896e855-73af-406e-9b99-8bfad958b99d%2Fj7on14a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 14: Pole Vaulting Physics**
A 75 kg pole vaulter running at 20 m/s vaults over the bar. If the vaulter's horizontal component of velocity over the bar is 15 m/s and air resistance is disregarded, how high was the jump?
**Explanation:**
To solve this problem, we need to understand the conservation of energy and components of velocity involved:
- **Initial Kinetic Energy (KE):**
\[
KE_{\text{initial}} = \frac{1}{2} m v_{\text{initial}}^2
\]
where \( m = 75 \, \text{kg} \) and the initial running speed \( v_{\text{initial}} = 20 \, \text{m/s} \).
- **Final Kinetic Energy (Horizontal Component):**
\[
KE_{\text{horizontal}} = \frac{1}{2} m v_{\text{horizontal}}^2
\]
where \( v_{\text{horizontal}} = 15 \, \text{m/s} \).
- **Potential Energy (PE) at the top of the jump:**
\[
PE = mgh
\]
where \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity), and \( h \) is the height to be determined.
**Conservation of Energy:**
The initial kinetic energy is converted into potential energy at the top of the vault and the kinetic energy of the horizontal motion:
\[
\frac{1}{2} m v_{\text{initial}}^2 = \frac{1}{2} m v_{\text{horizontal}}^2 + mgh
\]
**Solving for \( h \):**
Rearrange the equation to find the height \( h \):
\[
h = \frac{\frac{1}{2} m v_{\text{initial}}^2 - \frac{1}{2} m v_{\text{horizontal}}^2}{mg}
\]
Simplify:
\[
h = \frac{\frac{1}{2} (v_{\text{initial}}^2 - v_{\text{horizontal}}^2)}{g}
\]
Substitute the known values:
\[
h = \frac{\frac{1}{2} (20^2 - 15^
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