(a) A 70-kg man steps out a window and falls (from rest) 1.2 m to a sidewalk. What is his speed just before his feet strike the pavement? m/s (b) If the man falls with his knees and ankles locked, the only cushion for his fall is an approximately 0.47-cm give in the pads of his feet. Calculate the average force exerted on him by the ground during this 0.47 cm of travel. This average force is tufficient to cause damage to cartilage in the joints or to break bones.
(a) A 70-kg man steps out a window and falls (from rest) 1.2 m to a sidewalk. What is his speed just before his feet strike the pavement? m/s (b) If the man falls with his knees and ankles locked, the only cushion for his fall is an approximately 0.47-cm give in the pads of his feet. Calculate the average force exerted on him by the ground during this 0.47 cm of travel. This average force is tufficient to cause damage to cartilage in the joints or to break bones.
College Physics
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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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![### Physics Problem: Falling Object
#### Problem Statement
**(a)**
A 70-kg man steps out a window and falls (from rest) 1.2 m to a sidewalk. What is his speed just before his feet strike the pavement?
\[ \text{Speed:} \quad \boxed{\text{ }} \text{m/s} \]
**(b)**
If the man falls with his knees and ankles locked, the only cushion for his fall is an approximately 0.47-cm give in the pads of his feet. Calculate the average force exerted on him by the ground during this 0.47 cm of travel. This average force is sufficient to cause damage to cartilage in the joints or to break bones.
\[ \text{Average Force:} \quad \boxed{\text{ }} \text{N} \]
---
### Detailed Explanation:
#### Part (a):
To determine the man's speed just before striking the pavement, we can use the kinematic equation for an object in free fall:
\[ v^2 = u^2 + 2gh \]
Where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity (which is 0 m/s in this case, as he starts from rest),
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²),
- \( h \) is the height of the fall (1.2 m).
So, we substitute the given values:
\[ v^2 = 0 + 2 \times 9.81 \times 1.2 \]
\[ v^2 = 23.544 \]
\[ v = \sqrt{23.544} \]
\[ v \approx 4.85 \text{ m/s} \]
#### Part (b):
First, we need to calculate the deceleration experienced by the man when he hits the ground. Using the kinematic equation:
\[ v_f^2 = v_i^2 + 2ad \]
Where:
- \( v_f = 0 \text{ m/s} \) (final velocity, as he comes to a stop),
- \( v_i = 4.85 \text{ m/s} \) (initial velocity, as calculated in part a),
- \( d = 0.47 \text{ cm} = 0.0047 \text{ m} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F32cd56cf-4bd6-4dcd-a322-7126cf104832%2Fc51185fd-affa-494c-8310-0456cfe5ef18%2F6c7kf1.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics Problem: Falling Object
#### Problem Statement
**(a)**
A 70-kg man steps out a window and falls (from rest) 1.2 m to a sidewalk. What is his speed just before his feet strike the pavement?
\[ \text{Speed:} \quad \boxed{\text{ }} \text{m/s} \]
**(b)**
If the man falls with his knees and ankles locked, the only cushion for his fall is an approximately 0.47-cm give in the pads of his feet. Calculate the average force exerted on him by the ground during this 0.47 cm of travel. This average force is sufficient to cause damage to cartilage in the joints or to break bones.
\[ \text{Average Force:} \quad \boxed{\text{ }} \text{N} \]
---
### Detailed Explanation:
#### Part (a):
To determine the man's speed just before striking the pavement, we can use the kinematic equation for an object in free fall:
\[ v^2 = u^2 + 2gh \]
Where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity (which is 0 m/s in this case, as he starts from rest),
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²),
- \( h \) is the height of the fall (1.2 m).
So, we substitute the given values:
\[ v^2 = 0 + 2 \times 9.81 \times 1.2 \]
\[ v^2 = 23.544 \]
\[ v = \sqrt{23.544} \]
\[ v \approx 4.85 \text{ m/s} \]
#### Part (b):
First, we need to calculate the deceleration experienced by the man when he hits the ground. Using the kinematic equation:
\[ v_f^2 = v_i^2 + 2ad \]
Where:
- \( v_f = 0 \text{ m/s} \) (final velocity, as he comes to a stop),
- \( v_i = 4.85 \text{ m/s} \) (initial velocity, as calculated in part a),
- \( d = 0.47 \text{ cm} = 0.0047 \text{ m} \
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