A ball is launched at an angle of 38 degrees. The ball is thrown at a height of 2.0 m and lands on a 2.0 m tall ledge. The ledge is 2.7 E 1 m away from the starting point. Determine the maximum height that the ball reaches above the ground.

To determine the maximum height that the ball reaches above the ground, we need to break down the motion into its horizontal and vertical components. We will follow these steps:

### 1. **Horizontal and Vertical Components of Initial Velocity**

The launch angle is \( \theta = 38^\circ \), and we denote the initial velocity of the ball as \( v_0 \). We can break down \( v_0 \) into horizontal and vertical components using trigonometry:

\[
v_{0x} = v_0 \cos(\theta)
\]
\[
v_{0y} = v_0 \sin(\theta)
\]

### 2. **Horizontal Motion**

The ball travels horizontally to the ledge, which is 27 m away. The horizontal distance traveled can be expressed as:

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

Where \( t \) is the time of flight. From this equation, we can solve for \( t \):

\[
t = \frac{x}{v_{0x}} = \frac{27}{v_0 \cos(38^\circ)}
\]

### 3. **Vertical Motion**

The ball is launched from a height of 2.0 m, and the ledge is also 2.0 m high, so the total vertical displacement is zero (\( y = 0 \)). The vertical position at any time is described by the equation:

\[
y = y_0 + v_{0y} t - \frac{1}{2} g t^2
\]

Where \( y_0 = 2.0 \, \text{m} \) is the initial height, and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

Since the ball lands at the same height from which it was thrown (\( y = y_0 \)), the vertical displacement at the final position is zero. So:

\[
0 = 2.0 + v_0 \sin(38^\circ) t - \frac{1}{2} g t^2
\]

### 4. **Determine Initial Velocity Using Time of Flight**

Next, we will solve the time of flight equation using the known horizontal distance and the equation of motion to find \( v_0 \). After that, we can determine the time when the ball reaches its maximum height and calculate that height.

I'll compute the intermediate steps to determine \( v_0 \) and the maximum height.The initial velocity of the ball is approximately \( v_0 = 15.78 \, \text{m/s} \).

### 5. **Maximum Height**

The ball reaches its maximum height when its vertical velocity becomes zero. The time to reach maximum height, \( t_{\text{max}} \), can be found using the vertical component of the velocity:

\[
v_{y} = v_0 \sin(\theta) - g t_{\text{max}} = 0
\]

Solving for \( t_{\text{max}} \):

\[
t_{\text{max}} = \frac{v_0 \sin(\theta)}{g}
\]

Now, the maximum height can be found by substituting \( t_{\text{max}} \) into the vertical position equation:

\[
y_{\text{max}} = y_0 + v_0 \sin(\theta) t_{\text{max}} - \frac{1}{2} g t_{\text{max}}^2
\]

I'll calculate this next.The maximum height that the ball reaches above the ground is approximately \( 6.82 \, \text{m} \).

Would you like more details on any part of the calculation? Here are 5 related questions for further exploration:

1. How does changing the launch angle affect the maximum height?
2. What would the flight time be if the ledge were higher or lower?
3. How does air resistance impact projectile motion in real-world scenarios?
4. What is the total time the ball is in the air in this situation?
5. How could you calculate the horizontal velocity of the ball at the moment of landing?

**Tip:** The maximum height in projectile motion always occurs when the vertical component of the velocity becomes zero.

Explore the physics of projectile motion by determining the maximum height of a ball launched at a 38-degree angle from 2.0 m above the ground. This solution uses trigonometry and kinematic equations to find the maximum height, suitable for high school students in Grades 10-12.

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