A ball is projected vertically down with an initial velocity from a height of 20m on to a horizontal floor. During the impact it loses 50% of its energy and rebounds to the same height. The velocity of projection is `(g=10ms^(-2))`

Options

A. `20ms^(-1)`
B. `15ms^(-1)`
C. `10ms^(-1)`
D. `5ms^(-1)`

Answer

To solve the problem step by step, we will analyze the energy of the ball before and after the impact.

### Step 1: Understand the Initial Energy
The ball is projected from a height of 20 m with an initial velocity \( u \). The total initial energy \( E_i \) of the ball is the sum of its kinetic energy (due to its initial velocity) and its potential energy (due to its height).

\[ E_i = \text{Kinetic Energy} + \text{Potential Energy} \]
\[ E_i = \frac{1}{2} m u^2 + mgh \]

Where:
- \( m \) is the mass of the ball
- \( g \) is the acceleration due to gravity (given as \( 10 \, \text{m/s}^2 \))
- \( h \) is the height (given as \( 20 \, \text{m} \))

### Step 2: Understand the Energy After Impact
During the impact, the ball loses 50% of its energy. Therefore, the energy after impact \( E_f \) is:

\[ E_f = \frac{1}{2} E_i \]

After the impact, the ball rebounds to the same height of 20 m, so the potential energy at the maximum height after rebounding is:

\[ E_f = mgh \]

### Step 3: Set Up the Equation
Since the energy after impact is equal to the potential energy at the maximum height:

\[ mgh = \frac{1}{2} \left( \frac{1}{2} m u^2 + mgh \right) \]

### Step 4: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):

\[ gh = \frac{1}{4} u^2 + \frac{1}{2} gh \]

Now, rearranging gives:

\[ gh - \frac{1}{2} gh = \frac{1}{4} u^2 \]

\[ \frac{1}{2} gh = \frac{1}{4} u^2 \]

### Step 5: Solve for \( u^2 \)
Multiply both sides by 4 to eliminate the fraction:

\[ 2gh = u^2 \]

### Step 6: Substitute Known Values
Now substitute \( g = 10 \, \text{m/s}^2 \) and \( h = 20 \, \text{m} \):

\[ u^2 = 2 \times 10 \times 20 \]
\[ u^2 = 400 \]

### Step 7: Solve for \( u \)
Taking the square root of both sides:

\[ u = \sqrt{400} = 20 \, \text{m/s} \]

### Final Answer
The velocity of projection \( u \) is \( 20 \, \text{m/s} \).

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