A ball is dropped from a height of 100 m. At the surface of the earth, 20% of its energy is lost. To what height the ball will rise?

Options

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

Explanation

To solve the problem of how high the ball will rise after being dropped from a height of 100 m and losing 20% of its energy upon impact, we can follow these steps:

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

Thus, the initial potential energy at height \( h \) is:
\[ PE = m \cdot g \cdot 100 \]

### Step 2: Determine the energy after the ball hits the ground
Since 20% of the energy is lost when the ball hits the ground, 80% of the initial energy is retained. Therefore, the energy remaining after the impact is:
\[ \text{Remaining Energy} = 0.8 \cdot PE = 0.8 \cdot (m \cdot g \cdot 100) \]

### Step 3: Set the remaining energy equal to the potential energy at the new height
When the ball rises again, the remaining energy will convert back into potential energy at the new height \( h' \). Thus, we can write:
\[ 0.8 \cdot (m \cdot g \cdot 100) = m \cdot g \cdot h' \]

### Step 4: Cancel out the mass and gravitational constant
Since \( m \) and \( g \) appear on both sides of the equation, we can cancel them out:
\[ 0.8 \cdot 100 = h' \]

### Step 5: Calculate the new height
Now, we can calculate \( h' \):
\[ h' = 0.8 \cdot 100 = 80 \, \text{m} \]

### Conclusion
The ball will rise to a height of **80 meters** after bouncing back from the ground.

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