A ball falls from a height of 10 m. On rebounding, it loses 30% energy. The ball goes upto a height of

Options

A. 5 m
B. 7 m
C. 6 m
D. 8 m

Explanation

To solve the problem of a ball falling from a height of 10 m and rebounding with a loss of 30% energy, we can follow these steps:

### Step 1: Determine the initial potential energy (PE) of the ball
The potential energy (PE) of the ball at the height of 10 m can be calculated using the formula:
\[ \text{PE} = mgh \]
Where:
- \( m \) = mass of the ball (we will see that it cancels out later)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height (10 m)

Thus, the initial potential energy at height \( h \) is:
\[ \text{PE} = mg \times 10 \]

### Step 2: Calculate the total energy just before the ball hits the ground
Just before hitting the ground, all the potential energy will have converted into kinetic energy (KE). Therefore, the kinetic energy just before impact is:
\[ \text{KE} = \text{PE} = mg \times 10 \]

### Step 3: Determine the energy after the rebound
The problem states that the ball loses 30% of its energy upon rebounding. Therefore, the energy retained after the rebound is:
\[ \text{Remaining Energy} = \text{Total Energy} - \text{Energy Lost} \]
\[ \text{Remaining Energy} = \text{KE} \times (1 - 0.3) = mg \times 10 \times 0.7 = 0.7 mg \times 10 \]

### Step 4: Calculate the height the ball will reach after rebounding
When the ball rebounds, it will convert the remaining energy back into potential energy at the new height \( h' \):
\[ \text{PE}_{\text{new}} = mg \times h' \]
Setting the remaining energy equal to the new potential energy gives:
\[ 0.7 mg \times 10 = mg \times h' \]

### Step 5: Simplify the equation
We can cancel \( mg \) from both sides (since it is not zero):
\[ 0.7 \times 10 = h' \]
\[ h' = 7 \, \text{m} \]

### Conclusion
The ball will rebound to a height of 7 m after losing 30% of its energy.

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