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The problem involves calculating the peak height \( h_C \) of a ball after its first bounce, given certain initial conditions. Let me walk through the steps.

### Given:
- \( m_{\text{ball}} \): Mass of the ball (in kg)
- \( h_A \): Initial height from which the ball is launched (in meters)
- \( V_A \): Initial velocity of the ball (in m/s)
- \( \theta_A \): Angle of launch with respect to the horizontal (in degrees)
- \( e \): Coefficient of restitution (dimensionless)
- \( h_B = 0 \): The ball hits the ground at point B

The coefficient of restitution \( e \) is a measure of how elastic the collision is between the ball and the ground. It is defined as the ratio of the ball's velocity after the bounce to its velocity just before the bounce, in the vertical direction:
\[
e = \frac{V_{C_y}}{V_{B_y}}
\]
where \( V_{B_y} \) is the vertical component of velocity just before the bounce, and \( V_{C_y} \) is the vertical component of velocity just after the bounce.

### Step-by-Step Approach:

1. **Determine Vertical and Horizontal Components of Initial Velocity**:
   The initial velocity \( V_A \) can be decomposed into horizontal and vertical components:
   \[
   V_{A_x} = V_A \cos(\theta_A)
   \]
   \[
   V_{A_y} = V_A \sin(\theta_A)
   \]

2. **Vertical Velocity at Point B**:
   To find the vertical velocity \( V_{B_y} \) just before the ball hits the ground, use the kinematic equation for vertical motion:
   \[
   V_{B_y}^2 = V_{A_y}^2 + 2g h_A
   \]
   where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

3. **Vertical Velocity After the Bounce**:
   Using the coefficient of restitution \( e \), we know the relationship between the vertical velocities before and after the bounce:
   \[
   V_{C_y} = e V_{B_y}
   \]
   
4. **Determine Peak Height After Bounce \( h_C \)**:
   The maximum height \( h_C \) after the bounce can be found using energy conservation or kinematic equations. The vertical velocity at the peak will be zero, and we can use:
   \[
   0 = V_{C_y}^2 - 2g h_C
   \]
   Solving for \( h_C \):
   \[
   h_C = \frac{V_{C_y}^2}{2g} = \frac{e^2 V_{B_y}^2}{2g}
   \]

Now that we have the general approach, you can plug in the values for \( V_A \), \( \theta_A \), \( h_A \), and \( e \) to get the specific value of \( h_C \).

Would you like further clarifications or need help with specific values? Let me know!

### Relative questions:
1. How does the coefficient of restitution affect the peak height after each bounce?
2. What would happen to the horizontal component of velocity during the bounce?
3. How can the launch angle \( \theta_A \) affect the horizontal distance between points A, B, and C?
4. What assumptions are made in this calculation about air resistance?
5. How would adding friction between the ball and the ground change the results?

**Tip**: Always check the unit consistency in physics problems to avoid calculation errors, especially with angles (degrees vs. radians) and velocities (component-wise).

A ball with mass of m_ball [kg] is launched from height h_A [m] at velocity V_A [m/s] and angle relative to horizontal at θ_A [deg]. If the coefficient of restitution between the ground and the ball is e [dimensionless], calculate h_C [m], the peak height of the ball after the first bounce. Consider the height of the ground such that h_B = 0.

This problem involves calculating the peak height of a ball after its first bounce using projectile motion principles and the coefficient of restitution. The ball is launched with velocity V_A at angle θ_A from a height h_A. Learn how to calculate the peak height h_C after the first bounce using key kinematic equations.

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