An inelastic ball is dropped from a height 100 meter. If due to impact it loses 35% of its energy the ball will rise to a height of

Options

A. 35 m
B. 65 m
C. 100 m
D. 135 m

Answer

To solve the problem step by step, we will follow the principles of energy conservation and the effects of energy loss during an inelastic collision.

### Step 1: Calculate the initial potential energy (PE) of the ball
The potential energy of an object at height \( h \) is given by the formula:
\[ PE = mgh \]
where:
- \( m \) = mass of the ball (we will keep it as \( m \) since it will cancel out later),
- \( g \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( h \) = height from which the ball is dropped (100 m).

Substituting the values:
\[ PE = mg \cdot 100 \]

### Step 2: Calculate the kinetic energy (KE) just before impact
When the ball is dropped from a height of 100 m, just before impact, all the potential energy will convert into kinetic energy. Thus:
\[ KE = PE = mg \cdot 100 \]

### Step 3: Determine the energy lost during the impact
The ball loses 35% of its energy upon impact. Therefore, the energy retained after the impact is:
\[ \text{Energy retained} = KE \times (1 - 0.35) = KE \times 0.65 \]
Substituting for \( KE \):
\[ \text{Energy retained} = mg \cdot 100 \times 0.65 \]

### Step 4: Calculate the height the ball will rise to after the impact
After the impact, the retained energy will convert back into potential energy as the ball rises. The potential energy at the maximum height \( h' \) is given by:
\[ PE' = mgh' \]
Setting the retained energy equal to the potential energy at the new height:
\[ mg \cdot 100 \times 0.65 = mgh' \]
Cancelling \( m \) from both sides and substituting \( g \):
\[ 100 \times 0.65 = h' \]
\[ h' = 65 \, \text{m} \]

### Conclusion
The ball will rise to a height of **65 meters** after the impact.

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