A ball is dropped from height h on horizontal floor. If it loses `60%` of its energy on hitting the floor then height upto which it will rise after first rebounce is

Options

A. `(h)/(5)`
B. `(2h)/(5)`
C. `(3h)/(5)`
D. `(4h)/(5)`

Answer

To solve the problem step by step, we will analyze the situation where a ball is dropped from a height \( h \) and loses 60% of its energy upon hitting the ground. We want to find out the height \( h' \) to which the ball will rise after the first rebound.

### Step 1: Determine the initial potential energy
When the ball is at height \( h \), its potential energy (PE) is given by:
\[ PE = mgh \]
where \( m \) is the mass of the ball and \( g \) is the acceleration due to gravity.

### Step 2: Calculate the kinetic energy just before impact
As the ball falls, its potential energy converts to kinetic energy (KE) just before it hits the ground. The kinetic energy just before impact is equal to the initial potential energy:
\[ KE_{initial} = mgh \]

### Step 3: Determine the energy lost upon impact
The problem states that the ball loses 60% of its kinetic energy upon hitting the floor. Therefore, the energy lost is:
\[ \text{Energy lost} = 0.6 \times KE_{initial} = 0.6 \times mgh \]

### Step 4: Calculate the remaining kinetic energy after impact
The remaining kinetic energy after the impact is:
\[ KE_{remaining} = KE_{initial} - \text{Energy lost} = mgh - 0.6 \times mgh = 0.4 \times mgh \]

### Step 5: Relate remaining kinetic energy to height after rebound
When the ball rebounds, the remaining kinetic energy is converted back into potential energy at the maximum height \( h' \):
\[ KE_{remaining} = PE_{final} = mgh' \]
Thus, we have:
\[ 0.4 \times mgh = mgh' \]

### Step 6: Solve for the height \( h' \)
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 0.4gh = gh' \]
Now, divide both sides by \( g \):
\[ 0.4h = h' \]
Thus, we find:
\[ h' = 0.4h \]

### Conclusion
The height to which the ball will rise after the first rebound is:
\[ h' = \frac{2}{5}h \]