A body falls on a surface of coefficient of restitution 0.6 from a height of 1 m . Then the body rebounds to a height of

Options

A. 0.6 m
B. 0.4 m
C. 1 m
D. 0.36 m

Answer

To solve the problem of a body falling from a height of 1 meter and rebounding with a coefficient of restitution of 0.6, we can follow these steps:

### Step 1: Calculate the velocity just before impact
Using the equation of motion:
\[ v^2 = u^2 + 2gs \]
where:
- \( u = 0 \) (initial velocity),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( s = 1 \, \text{m} \) (height fallen).

Substituting the values:
\[ v^2 = 0 + 2 \times 10 \times 1 \]
\[ v^2 = 20 \]
\[ v = \sqrt{20} \]
\[ v \approx 4.47 \, \text{m/s} \]

### Step 2: Calculate the velocity after the collision
The coefficient of restitution \( e \) is defined as:
\[ e = \frac{\text{velocity after collision}}{\text{velocity before collision}} \]
Given \( e = 0.6 \):
\[ \text{velocity after collision} = e \times \text{velocity before collision} \]
\[ \text{velocity after collision} = 0.6 \times 4.47 \]
\[ \text{velocity after collision} \approx 2.68 \, \text{m/s} \]

### Step 3: Calculate the maximum height after rebounding
Using the same equation of motion:
\[ v^2 = u^2 - 2gh \]
At the maximum height, the final velocity \( v = 0 \):
\[ 0 = u^2 - 2gh \]
Rearranging gives:
\[ h = \frac{u^2}{2g} \]
Substituting \( u = 2.68 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ h = \frac{(2.68)^2}{2 \times 10} \]
\[ h = \frac{7.1824}{20} \]
\[ h \approx 0.359 \, \text{m} \]

### Step 4: Convert height to centimeters
To convert meters to centimeters:
\[ h \approx 0.359 \, \text{m} \times 100 \]
\[ h \approx 35.9 \, \text{cm} \]

### Final Answer
The body rebounds to a height of approximately **35.9 cm**.

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