To solve the problem, we will use the principles of energy conservation and kinematics. Let's break it down step by step.
### Step 1: Understand the Energy at the Initial Position
The ball is thrown from a height of 20 m with an initial velocity \( v_0 \). At this point, the total mechanical energy (E_initial) of the ball consists of its potential energy (PE) and kinetic energy (KE).
\[
E_{\text{initial}} = PE + KE = mgh + \frac{1}{2}mv_0^2
\]
Where:
- \( m \) is the mass of the ball (which will cancel out later),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 20 \, \text{m} \) (initial height).
### Step 2: Energy Just Before Impact
When the ball reaches the ground, its height is 0, and all of its energy is converted into kinetic energy (just before impact). The potential energy at the ground is zero.
\[
E_{\text{just before impact}} = \frac{1}{2}mv^2
\]
Where \( v \) is the velocity just before impact.
### Step 3: Energy After Collision
Upon colliding with the ground, the ball loses 50% of its energy. Therefore, the energy after the collision (E_final) is half of the energy just before impact.
\[
E_{\text{final}} = \frac{1}{2} \left( \frac{1}{2}mv^2 \right) = \frac{1}{4}mv^2
\]
### Step 4: Energy at Maximum Height After Rebound
After rebounding, the ball reaches the same height of 20 m again. At this height, all of the energy is potential energy.
\[
E_{\text{final}} = mgh
\]
### Step 5: Set Up the Equation
Now, we can equate the energy after the collision to the potential energy at the maximum height after rebounding:
\[
\frac{1}{4}mv^2 = mgh
\]
Since \( m \) appears on both sides, we can cancel it out:
\[
\frac{1}{4}v^2 = gh
\]
### Step 6: Substitute for \( v \)
We know that just before impact, the velocity \( v \) can be found using energy conservation from the initial position to just before impact:
\[
mgh + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2
\]
This simplifies to:
\[
gh + \frac{1}{2}v_0^2 = \frac{1}{2}v^2
\]
### Step 7: Solve for \( v^2 \)
Rearranging gives:
\[
v^2 = 2gh + v_0^2
\]
### Step 8: Substitute \( v^2 \) into the Energy Equation
Now substitute \( v^2 \) back into the energy equation:
\[
\frac{1}{4}(2gh + v_0^2) = gh
\]
### Step 9: Solve for \( v_0^2 \)
Multiply through by 4 to eliminate the fraction:
\[
2gh + v_0^2 = 4gh
\]
Rearranging gives:
\[
v_0^2 = 4gh - 2gh = 2gh
\]
### Step 10: Substitute Values
Now substitute \( g = 10 \, \text{m/s}^2 \) and \( h = 20 \, \text{m} \):
\[
v_0^2 = 2 \times 10 \times 20 = 400
\]
### Step 11: Calculate \( v_0 \)
Taking the square root gives:
\[
v_0 = \sqrt{400} = 20 \, \text{m/s}
\]
### Final Answer
The initial velocity \( v_0 \) is \( 20 \, \text{m/s} \).
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