An inelastic ball is dropped from a height of 100 m. Due to earth 20% of its energy is lost. To what height the ball will rise ?

Options

A. 80 m
B. 40 m
C. 60 m
D. 20 m

Explanation

To solve the problem step by step, we will analyze the situation using the principles of energy conservation.

### Step 1: Calculate the initial potential energy (PE_initial)
The ball is dropped from a height of 100 m. The potential energy (PE) at this height can be calculated using the formula:
\[ PE_{\text{initial}} = mgh \]
where:
- \( m \) is the mass of the ball (we can keep it as \( m \) since it will cancel out later),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height (100 m).

Thus,
\[ PE_{\text{initial}} = mg \times 100 \]

### Step 2: Calculate the energy lost upon impact
It is given that 20% of the energy is lost when the ball hits the ground. Therefore, the energy retained after the impact is:
\[ \text{Energy retained} = PE_{\text{initial}} - \text{Energy lost} \]
\[ \text{Energy lost} = 0.20 \times PE_{\text{initial}} = 0.20 \times (mg \times 100) = 20mg \]
So,
\[ \text{Energy retained} = PE_{\text{initial}} - 20mg = mg \times 100 - 20mg = 80mg \]

### Step 3: Relate the retained energy to the height the ball will rise
After the impact, the retained energy will be converted into potential energy as the ball rises. The potential energy at the new height \( h_1 \) can be expressed as:
\[ PE_{\text{final}} = mgh_1 \]
Setting the retained energy equal to the potential energy at the new height:
\[ 80mg = mgh_1 \]

### Step 4: Solve for the height \( h_1 \)
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 80g = gh_1 \]
Dividing both sides by \( g \):
\[ h_1 = 80 \, \text{m} \]

### Conclusion
The ball will rise to a height of **80 meters** after the impact.

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