After falling from a height h and striking the ground twice, a ball rise up to the height [e = coefficient of restitution]

Options

A. he
B. `he^(2)`
C. `he^(3)`
D. `he^(4)`

Answer

To solve the problem of a ball falling from a height \( h \) and striking the ground twice before rising to a new height, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Coefficient of Restitution (e)**:
The coefficient of restitution \( e \) is defined as the ratio of the relative velocity of separation to the relative velocity of approach. For a ball dropping and bouncing back, it relates the velocities before and after the collision with the ground.

2. **Calculating the Velocity Just Before the First Impact**:
When the ball falls from a height \( h \), we can use the equation of motion to find its velocity just before it strikes the ground for the first time. The equation is:
\[ v = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity.

3. **Calculating the Velocity After the First Impact**:
After the first impact, the velocity of the ball when it rebounds is given by:
\[ v' = e \cdot v = e \cdot \sqrt{2gh} \]

4. **Calculating the Height After the First Bounce**:
The height \( h_1 \) that the ball reaches after the first bounce can be calculated using the kinetic energy at the moment of impact:
\[ h_1 = \frac{(v')^2}{2g} = \frac{(e \cdot \sqrt{2gh})^2}{2g} = \frac{e^2 \cdot 2gh}{2g} = e^2 h \]

5. **Calculating the Velocity Just Before the Second Impact**:
The ball will fall from the height \( h_1 \) and just before the second impact, its velocity \( v'' \) will be:
\[ v'' = \sqrt{2gh_1} = \sqrt{2g(e^2 h)} = \sqrt{2gh} \cdot e \]

6. **Calculating the Velocity After the Second Impact**:
After the second impact, the velocity of the ball when it rebounds is:
\[ v''' = e \cdot v'' = e \cdot (\sqrt{2gh} \cdot e) = e^2 \cdot \sqrt{2gh} \]

7. **Calculating the Height After the Second Bounce**:
The height \( h_2 \) that the ball reaches after the second bounce can be calculated similarly:
\[ h_2 = \frac{(v''')^2}{2g} = \frac{(e^2 \cdot \sqrt{2gh})^2}{2g} = \frac{e^4 \cdot 2gh}{2g} = e^4 h \]

### Final Result:
After falling from a height \( h \) and striking the ground twice, the ball rises to a height of:
\[ h_2 = e^4 h \]