An archer shoots an arrow parallel to the ground. The arrow is found stuck in the ground 73 m away from where the archer fired, and it forms a 3-degree angle with the ground. What was the arrow’s initial velocity? 

To solve for the arrow's initial velocity, we can break the problem down into components: horizontal and vertical motion. Here’s how we approach it.

### Step 1: Determine the total time of flight (T)
The arrow hits the ground at an angle of 3° relative to the ground. This angle provides information about the relationship between the horizontal and vertical velocities at the moment of impact.

Let:
- \( \theta = 3^\circ \) be the angle of impact.
- \( v_x \) be the horizontal velocity.
- \( v_y \) be the vertical velocity at impact.

The relation between these velocities is given by:
\[
\tan(\theta) = \frac{v_y}{v_x}
\]
So,
\[
v_y = v_x \tan(3^\circ)
\]

### Step 2: Horizontal motion
In projectile motion, the horizontal velocity (\( v_x \)) remains constant throughout the flight because there is no acceleration in the horizontal direction (assuming no air resistance). The horizontal distance traveled is 73 m.

\[
d = v_x \cdot T
\]
where \( d = 73 \, \text{m} \) and \( T \) is the total time of flight. Therefore,
\[
T = \frac{d}{v_x}
\]

### Step 3: Vertical motion
The arrow’s vertical motion is influenced by gravity. The vertical velocity at the moment of impact (\( v_y \)) can be found using the kinematic equation:
\[
v_y = g \cdot T
\]
where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Now substitute \( T = \frac{d}{v_x} \) from the horizontal motion equation into this equation for vertical velocity:
\[
v_y = g \cdot \frac{d}{v_x}
\]

### Step 4: Combine vertical and horizontal components
From Step 1, we already know that \( v_y = v_x \cdot \tan(3^\circ) \). Now equate both expressions for \( v_y \):
\[
g \cdot \frac{d}{v_x} = v_x \cdot \tan(3^\circ)
\]

Now, solve for \( v_x \):
\[
g \cdot d = v_x^2 \cdot \tan(3^\circ)
\]
\[
v_x^2 = \frac{g \cdot d}{\tan(3^\circ)}
\]
\[
v_x = \sqrt{\frac{g \cdot d}{\tan(3^\circ)}}
\]

### Step 5: Calculate the horizontal velocity
Substitute the known values (\( g = 9.8 \, \text{m/s}^2 \), \( d = 73 \, \text{m} \), and \( \tan(3^\circ) \approx 0.05241 \)):
\[
v_x = \sqrt{\frac{9.8 \cdot 73}{0.05241}}
\]
Now, calculate this expression to find \( v_x \).

The initial horizontal velocity of the arrow is approximately \( 116.84 \, \text{m/s} \).

Would you like more details or have any questions? Here are some related questions to further explore this topic:

1. How would air resistance affect the arrow's flight and initial velocity?
2. What is the total time of flight for the arrow in this scenario?
3. How can we calculate the maximum height reached by the arrow during its flight?
4. If the angle with the ground was 5°, how would that change the initial velocity?
5. How would the solution change if the arrow was fired at an upward angle instead of parallel to the ground?

**Tip:** When dealing with projectile motion, always separate the horizontal and vertical components of the motion to simplify the problem.

An archer shoots an arrow parallel to the ground. The arrow is found stuck in the ground 73 m away from where the archer fired, and it forms a 3-degree angle with the ground. What was the arrow’s initial velocity?

Discover how to calculate the initial velocity of an arrow fired parallel to the ground, given a 3-degree angle upon impact and a distance of 73 meters. This problem applies concepts from kinematics and trigonometry, ideal for Grades 10-12.

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