A ball is dropped onto a floor from a height of `10 m`. If `20%` of its initial energy is lost,then the height of bounce is

Options

A. `2 m`
B. `4 m`
C. `8 m`
D. `6.4 m`

Explanation

To solve the problem of finding the height of the bounce of a ball dropped from a height of 10 meters, with 20% of its initial energy lost, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Potential Energy (PE_initial)**:
The initial potential energy (PE_initial) of the ball when dropped from a height (H) of 10 meters is given by the formula:
\[ PE_{initial} = mgh \]
where \( m \) is the mass of the ball, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \)), and \( h \) is the height (10 m). Thus, we have:
\[ PE_{initial} = mg \times 10 \]

2. **Calculate Energy Lost**:
According to the problem, 20% of the initial energy is lost when the ball hits the floor. Therefore, the energy lost (E_lost) can be calculated as:
\[ E_{lost} = 0.2 \times PE_{initial} = 0.2 \times (mg \times 10) = 2mg \]

3. **Determine Final Energy (PE_final)**:
The final energy (PE_final) after the ball has lost 20% of its energy is:
\[ PE_{final} = PE_{initial} - E_{lost} = (mg \times 10) - (2mg) = 8mg \]

4. **Relate Final Energy to Height of Bounce**:
The final energy when the ball bounces back up is converted into potential energy at the new height (h'). Thus, we can set the final energy equal to the potential energy at height h':
\[ PE_{final} = mgh' \]
Substituting the expression for PE_final:
\[ 8mg = mgh' \]

5. **Cancel Mass and Solve for h'**:
We can cancel \( m \) from both sides of the equation (assuming \( m \neq 0 \)):
\[ 8g = gh' \]
Dividing both sides by \( g \):
\[ 8 = h' \]
Therefore, the height of the bounce (h') is:
\[ h' = 8 \, \text{meters} \]

### Final Answer:
The height of the bounce is **8 meters**.

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