A batter hits a pitched ball when the center of the ball is  1.24m above the ground. The ball leaves the bat at an angle of 45 degrees with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of  107m. (a) Does the ball clear a 7.96 high fence that is 97m horizontally from the launch point? (b) At the fence, what is the distance between the fence top and the ball center?

### Problem Overview:

The batter hits the ball when it is 1.24 m above the ground, with a launch angle of 45° and an expected horizontal range of 107 m. The fence, which is 7.96 m tall, is located 97 m horizontally from the launch point. 

We need to determine two things:
1. **(a)** Does the ball clear the fence at 97 m horizontally?
2. **(b)** What is the vertical distance between the top of the fence and the ball when it reaches the fence?

### Step-by-Step Solution:

We'll start by calculating the necessary parameters of the ball's trajectory.

#### Given:
- Launch angle \( \theta = 45^\circ \)
- Horizontal range \( R = 107 \, \text{m} \)
- Initial height \( y_0 = 1.24 \, \text{m} \)
- Horizontal distance to fence \( x_f = 97 \, \text{m} \)
- Fence height \( y_f = 7.96 \, \text{m} \)

We will use the **equation for projectile motion** with an initial height \( y_0 \).

The horizontal and vertical components of the motion are governed by these equations:

1. **Horizontal motion:**
   \[
   x = v_0 \cos(\theta) \cdot t
   \]
   
2. **Vertical motion:**
   \[
   y = y_0 + v_0 \sin(\theta) \cdot t - \frac{1}{2} g t^2
   \]

Where:
- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.
- \( v_0 \) is the initial velocity of the ball.

#### Step 1: Find the initial velocity \( v_0 \)

The range formula for a projectile launched from an initial height is:
\[
R = \frac{v_0^2 \sin(2\theta)}{g} + \frac{2v_0 \sin(\theta) \cdot y_0}{g}
\]

Given \( R = 107 \, \text{m} \), \( y_0 = 1.24 \, \text{m} \), and \( \theta = 45^\circ \), let's solve for \( v_0 \). We'll then use this \( v_0 \) to find the vertical position of the ball when it reaches the fence.

It seems there is an issue with the initial velocity calculation, likely due to a misinterpretation of the range equation. Let's refine the approach, ensuring the formula is correctly applied, and re-calculate the initial velocity.The negative solution for initial velocity suggests an issue, possibly due to a sign convention mishandling. Let me adjust the approach by directly using time to solve for the motion equations and compute the ball's height when it reaches the fence.### Results:
1. **(a)** The ball clears the fence. At a horizontal distance of 97 m, the height of the ball is approximately **14.94 m**, which is much higher than the 7.96 m tall fence.
   
2. **(b)** The distance between the top of the fence and the ball's center is approximately **6.98 m**.

Would you like any additional details or clarifications on the solution?

Here are five related questions to further explore the topic:
1. How would the result change if the launch angle was 30° instead of 45°?
2. How does air resistance affect the ball's trajectory in real-world scenarios?
3. How can we calculate the total time of flight for the ball?
4. What is the ball's velocity when it reaches the fence?
5. How would the trajectory change if the batter hit the ball from a height of 1.5 m instead?

**Tip:** When solving projectile motion problems, breaking the motion into horizontal and vertical components simplifies the analysis.

A batter hits a pitched ball when the center of the ball is 1.24m above the ground. The ball leaves the bat at an angle of 45 degrees with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 107m. (a) Does the ball clear a 7.96m high fence that is 97m horizontally from the launch point? (b) At the fence, what is the distance between the fence top and the ball center?

This problem explores the trajectory of a ball hit at a 45-degree angle, covering kinematics, projectile motion, and trigonometry. Find out if the ball clears a 7.96m high fence that is 97m away. Step-by-step solution included, suitable for high school students (Grades 10-12).

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