https://file.aiquickdraw.com/mathbot/file/1729485987687x5xu53c2.jpg

Let's work through part (c) of the problem.

### Problem Setup:

A baseball of mass \( m = 0.19 \, \text{kg} \) is pushed onto a vertical spring that compresses a distance \( d = 0.18 \, \text{m} \). The spring constant is \( k = 630 \, \text{N/m} \), and gravitational potential energy is zero at the lowest position of the ball. From earlier parts of the problem:

- **Speed just after the ball leaves the spring**: \( v = 10.36 \, \text{m/s} \) (from part a).
- **Maximum height reached by the ball**: \( h = 5.30 \, \text{m} \) (from part b).

Now, we need to calculate the ball's velocity at **half the maximum height**.

### Key Steps:

1. **Energy Conservation**: We'll use conservation of mechanical energy (kinetic + potential energy) between the two points (at release and at half the height). 

   Total mechanical energy is conserved, so:
   \[
   E_{\text{total}} = E_{\text{kinetic}} + E_{\text{potential}}
   \]
   
   At maximum height (\( h = 5.30 \, \text{m} \)), all energy is potential, and the velocity is zero.
   
   At half the height (\( h/2 = 2.65 \, \text{m} \)), the energy is partly kinetic and partly potential.

2. **Mechanical Energy at Different Points**:
   - At \( h = 5.30 \, \text{m} \) (maximum height): All energy is gravitational potential.
     \[
     E_{\text{max}} = mgh = 0.19 \times 9.81 \times 5.30 = 9.86157 \, \text{J}
     \]
   - At \( h = 2.65 \, \text{m} \) (half the maximum height): 
     \[
     E_{\text{total}} = E_{\text{kinetic}} + E_{\text{potential}}
     \]
     The potential energy at half height is:
     \[
     U_{\text{half}} = mgh_{\text{half}} = 0.19 \times 9.81 \times 2.65 = 4.930785 \, \text{J}
     \]
     Therefore, the kinetic energy at half height is the remaining energy:
     \[
     K_{\text{half}} = E_{\text{total}} - U_{\text{half}} = 9.86157 - 4.930785 = 4.930785 \, \text{J}
     \]
     And from the kinetic energy formula, \( K = \frac{1}{2}mv^2 \), solve for \( v \):
     \[
     \frac{1}{2} \times 0.19 \times v^2 = 4.930785
     \]
     \[
     v^2 = \frac{4.930785 \times 2}{0.19} = 51.8
     \]
     \[
     v = \sqrt{51.8} \approx 7.2 \, \text{m/s}
     \]

### Answer:
The velocity at half the maximum height is approximately \( 7.2 \, \text{m/s} \).

Do you need further details or have any questions?

Here are 5 related questions for practice:
1. How would the velocity change if the spring constant were doubled?
2. What is the velocity at one-quarter the maximum height?
3. How would you calculate the velocity if there were air resistance?
4. How does changing the mass of the baseball affect the height it reaches?
5. What would happen to the mechanical energy if the equilibrium point of the spring were not at zero potential energy?

**Tip**: Always check if mechanical energy is conserved before applying energy equations.

What is the ball's velocity, in meters per second, at half of the maximum height relative to the equilibrium point of the spring?

Learn how to find the velocity of a baseball at half the maximum height using principles of energy conservation. This problem involves concepts like kinetic and potential energy, and is suitable for students in Grade 11-12 physics.

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