A ball is projected vertically down with an initial velocity from a height of `20 m` onto a horizontal floor. During the impact it loses `50%` of its energy and rebounds to the same height. The initial velocity of its projection is

Options

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

Explanation

To solve the problem, we need to find the initial velocity of a ball projected vertically downward from a height of 20 m, which loses 50% of its energy upon impact and rebounds to the same height.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Height (h) = 20 m
- Loss of energy during impact = 50%
- Gravitational acceleration (g) = 10 m/s² (assuming standard value)

2. **Calculate Total Energy at the Initial Height:**
- The total mechanical energy (E) at the height before projection consists of potential energy (PE) and kinetic energy (KE):
\[ E = PE + KE = mgh + \frac{1}{2}mv^2 \]
- Here, \(m\) is the mass of the ball, \(g\) is the acceleration due to gravity, and \(v\) is the initial velocity of projection.

3. **Energy After Impact:**
- After the impact, the ball loses 50% of its total energy. Therefore, the energy after impact is:
\[ E' = \frac{1}{2}E = \frac{1}{2} \left(mgh + \frac{1}{2}mv^2\right) \]

4. **Energy at Maximum Height After Rebounding:**
- The ball rebounds to the same height (20 m), so the potential energy at the maximum height after rebounding is:
\[ E' = mgh \]

5. **Set Up the Equation:**
- Since the energy after impact is equal to the potential energy at the maximum height:
\[ \frac{1}{2} \left(mgh + \frac{1}{2}mv^2\right) = mgh \]

6. **Simplify the Equation:**
- Cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ \frac{1}{2} \left(gh + \frac{1}{2}v^2\right) = gh \]
- Multiply through by 2:
\[ gh + \frac{1}{2}v^2 = 2gh \]
- Rearranging gives:
\[ \frac{1}{2}v^2 = 2gh - gh \]
\[ \frac{1}{2}v^2 = gh \]

7. **Solve for Initial Velocity (v):**
- Multiply both sides by 2:
\[ v^2 = 2gh \]
- Substitute \(g = 10 \, \text{m/s}^2\) and \(h = 20 \, \text{m}\):
\[ v^2 = 2 \times 10 \times 20 = 400 \]
- Taking the square root:
\[ v = \sqrt{400} = 20 \, \text{m/s} \]

### Final Answer:
The initial velocity of the ball projected downward is **20 m/s**.