A ball at rest is dropped freely from a height of `20 m`. It loses `30 %` of its energy on striking the ground and bounces back. The height to which it bounces back is

Options

A. 14 m
B. 12 m
C. 9 m
D. 6 m

Explanation

To solve the problem step by step, we need to follow these calculations:

### Step 1: Calculate the initial potential energy (PE) of the ball
The potential energy (PE) at a height \( h \) is given by the formula:
\[ PE = mgh \]
Where:
- \( m \) = mass of the ball (we can keep it as \( m \) since it will cancel out later)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height from which the ball is dropped (given as \( 20 \, m \))

So, the initial potential energy when the ball is at a height of \( 20 \, m \) is:
\[ PE = mg \times 20 \]

### Step 2: Calculate the energy lost on striking the ground
The ball loses \( 30\% \) of its energy upon striking the ground. Therefore, the energy retained after the impact is:
\[ \text{Energy retained} = (1 - 0.30) \times PE = 0.70 \times PE \]

### Step 3: Relate the retained energy to the height of the bounce
When the ball bounces back, all the retained kinetic energy converts back into potential energy at the maximum height it reaches. Let the height to which it bounces back be \( h_2 \). The potential energy at this height is:
\[ PE' = mgh_2 \]

Since the energy retained after the bounce is equal to the potential energy at the height \( h_2 \):
\[ 0.70 \times (mg \times 20) = mg \times h_2 \]

### Step 4: Cancel out the mass and gravitational acceleration
We can cancel \( m \) and \( g \) from both sides of the equation:
\[ 0.70 \times 20 = h_2 \]

### Step 5: Calculate the height \( h_2 \)
Now, calculate \( h_2 \):
\[ h_2 = 0.70 \times 20 = 14 \, m \]

Thus, the height to which the ball bounces back is **14 meters**.

### Final Answer:
The height to which the ball bounces back is **14 m**.

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