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Let's break down and solve this projectile motion problem step by step.

### Given:
- Initial velocity (\(v_0\)) = 25.0 m/s
- Launch angle (\(\theta_0\)) = 36.0° above the horizontal
- Horizontal distance to the wall (\(d\)) = 25.0 m

We need to find:
1. **(a)** The height at which the ball hits the wall, i.e., how far above the release point.
2. **(b)** The horizontal component of the velocity when the ball hits the wall.
3. **(c)** The vertical component of the velocity when the ball hits the wall.

### Step 1: Horizontal and Vertical Components of Initial Velocity

We start by resolving the initial velocity into horizontal and vertical components:

- Horizontal velocity: 
  \[
  v_{0x} = v_0 \cos(\theta_0) = 25.0 \cos(36.0^\circ)
  \]
  
- Vertical velocity:
  \[
  v_{0y} = v_0 \sin(\theta_0) = 25.0 \sin(36.0^\circ)
  \]

### Step 2: Time to Reach the Wall

To find the time \(t\) it takes for the ball to reach the wall, we use the fact that the horizontal velocity is constant in projectile motion. The formula for horizontal displacement is:

\[
d = v_{0x} t
\]

Solving for \(t\):

\[
t = \frac{d}{v_{0x}} = \frac{25.0}{v_{0x}}
\]

### Step 3: Height at which the ball hits the wall

Once we know the time, we can calculate how high the ball is at that moment. The vertical position of the ball as a function of time is given by:

\[
y = v_{0y} t - \frac{1}{2} g t^2
\]

where \(g = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity.

### Step 4: Velocity components at the wall

- **Horizontal velocity** remains constant, so it's simply \(v_{0x}\).
  
- **Vertical velocity** at the wall can be found using the equation:

\[
v_y = v_{0y} - g t
\]

Let me calculate the values based on these formulas.Here are the results:

1. **(a)** The ball hits the wall **10.68 m** above the release point.
2. **(b)** The horizontal component of the velocity when the ball hits the wall is **20.23 m/s**.
3. **(c)** The vertical component of the velocity when the ball hits the wall is **2.58 m/s**.

Would you like more details or clarification on any step?

Here are five related questions to extend your understanding:
1. How would the results change if the angle of projection were increased?
2. What is the total time of flight if the ball were to hit the ground after being thrown?
3. How far would the ball travel horizontally if it were thrown with the same initial velocity but no wall present?
4. How would air resistance affect the results of this problem?
5. What would happen if the wall were closer or farther from the release point?

**Tip**: In projectile motion, the horizontal velocity remains constant (in the absence of air resistance), while the vertical velocity changes due to gravity.

You throw a ball toward a wall at a speed of 25.0 m/s and at an angle of 36.0° above the horizontal. The wall is 25.0 m from the release point. How far above the release point does the ball hit the wall? What are the horizontal and vertical components of its velocity as it hits the wall?

Solve a projectile motion problem involving a ball thrown at a speed of 25.0 m/s and an angle of 36.0°. Learn how to calculate the height at which the ball hits the wall and the horizontal and vertical velocity components using trigonometry and kinematics. Suitable for high school physics students.

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