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. 1m
D. 0.36 m

Answer

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

### Step 1: Determine the velocity just before impact
When the body falls from a height \( h = 1 \, \text{m} \), we can use the formula for the velocity of a freely falling object just before impact:

\[ v = \sqrt{2gh} \]

Where:
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( h \) is the height (1 m).

Substituting the values:

\[ v = \sqrt{2 \times 10 \times 1} = \sqrt{20} \approx 4.47 \, \text{m/s} \]

### Step 2: Apply the coefficient of restitution
The coefficient of restitution \( e \) relates the velocities before and after the collision. It is given that \( e = 0.6 \). The relationship is:

\[ v' = e \cdot v \]

Where:
- \( v' \) is the velocity after the collision,
- \( v \) is the velocity before the collision.

Substituting the values:

\[ v' = 0.6 \times 4.47 \approx 2.68 \, \text{m/s} \]

### Step 3: Calculate the maximum height after rebound
Now, we need to find the maximum height \( h' \) that the body will reach after rebounding with the velocity \( v' \). We can use the formula:

\[ h' = \frac{(v')^2}{2g} \]

Substituting the values:

\[ h' = \frac{(2.68)^2}{2 \times 10} \]

Calculating \( (2.68)^2 \):

\[ (2.68)^2 \approx 7.1824 \]

Now substituting back into the height formula:

\[ h' = \frac{7.1824}{20} \approx 0.35912 \, \text{m} \]

### Step 4: Round off the answer
Rounding off to two decimal places, we find:

\[ h' \approx 0.36 \, \text{m} \]

Thus, the body rebounds to a height of approximately **0.36 meters**.

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