The hole on a level, elevated golf green is a horizontal distance of 120 meters from the tee and a vertical distance of 12.8 meters above the tee. A golfer hits a ball at an angle of 9.2 degrees and makes a hole in one. What was the initial speed of the ball?

We can solve this problem by analyzing the horizontal and vertical components of the ball's motion using kinematic equations. Let's break it down:

### Given:
- Horizontal distance to the hole, \( d_x = 120 \, \text{m} \)
- Vertical height difference, \( h = 12.8 \, \text{m} \)
- Launch angle, \( \theta = 9.2^\circ \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
- The ball lands in the hole, which means its final position matches the target.

We need to find the initial speed of the ball, \( v_0 \).

### Step 1: Horizontal motion equation
The horizontal distance covered is given by:
\[
d_x = v_{0x} \cdot t
\]
where \( v_{0x} = v_0 \cdot \cos(\theta) \) is the horizontal component of the initial velocity, and \( t \) is the time of flight.

\[
t = \frac{d_x}{v_0 \cdot \cos(\theta)}
\]

### Step 2: Vertical motion equation
For vertical motion, we use the equation:
\[
y = v_{0y} \cdot t - \frac{1}{2} g t^2
\]
where \( y = h \), and \( v_{0y} = v_0 \cdot \sin(\theta) \) is the vertical component of the initial velocity.

Substituting the expression for \( t \) from the horizontal motion equation into the vertical motion equation:
\[
h = v_0 \cdot \sin(\theta) \cdot \frac{d_x}{v_0 \cdot \cos(\theta)} - \frac{1}{2} g \left( \frac{d_x}{v_0 \cdot \cos(\theta)} \right)^2
\]

Simplifying the equation:
\[
h = d_x \cdot \tan(\theta) - \frac{g d_x^2}{2 v_0^2 \cdot \cos^2(\theta)}
\]

### Step 3: Solve for \( v_0 \)
Now we can solve for \( v_0 \). Rearranging the above equation:

\[
h = d_x \cdot \tan(\theta) - \frac{g d_x^2}{2 v_0^2 \cdot \cos^2(\theta)}
\]
\[
\frac{g d_x^2}{2 v_0^2 \cdot \cos^2(\theta)} = d_x \cdot \tan(\theta) - h
\]
\[
v_0^2 = \frac{g d_x^2}{2 \cdot \cos^2(\theta) \cdot (d_x \cdot \tan(\theta) - h)}
\]

Taking the square root:
\[
v_0 = \sqrt{\frac{g d_x^2}{2 \cdot \cos^2(\theta) \cdot (d_x \cdot \tan(\theta) - h)}}
\]

Now, let's plug in the values:

- \( g = 9.8 \, \text{m/s}^2 \)
- \( d_x = 120 \, \text{m} \)
- \( \theta = 9.2^\circ \) (convert to radians: \( \theta = 9.2 \times \frac{\pi}{180} \))
- \( h = 12.8 \, \text{m} \)

I'll calculate the value of \( v_0 \).The initial speed of the ball was approximately \( 104.46 \, \text{m/s} \).

Would you like a detailed explanation of any part of the solution?

Here are five related questions that could deepen your understanding:
1. How do the horizontal and vertical components of velocity affect the trajectory?
2. What happens to the initial speed if the launch angle increases?
3. How would air resistance alter the calculations for a golf ball?
4. What would be the effect of changing the elevation of the golf hole?
5. Can you derive the formula for the time of flight in projectile motion?

**Tip:** In projectile motion problems, always break the motion into horizontal and vertical components for easier analysis.

Learn how to calculate the initial speed of a golf ball hit at an angle of 9.2 degrees to make a hole in one, covering a horizontal distance of 120 meters and a vertical elevation of 12.8 meters. Explore key concepts like projectile motion, kinematic equations, and trigonometric functions. Suitable for undergraduate physics students.

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