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Let's analyze the given problem step by step.

### Problem Overview:

- **Mass of the fragment**: \(2.0 \, \text{kg}\)
- **Distance traveled North**: \(103 \, \text{m}\)
- **Final position height**: \(11.0 \, \text{m}\) above ground
- **Initial height**: \(0.45 \, \text{m}\) above ground
- **Impact speed**: \(40.0 \, \text{m/s}\) at an angle \(12.9^\circ\) below the horizontal
- **Fragment's trajectory**: It traveled without hitting anything else in the way.

### Objective:
1. **Find the magnitude of the initial velocity.**
2. **Determine the angle the initial velocity makes with the horizontal.**

### Step 1: Vertical motion analysis
The vertical component of the velocity can be analyzed using the following kinematic equation for displacement:
\[
y_f = y_0 + v_{y0}t + \frac{1}{2}at^2
\]
where:
- \(y_f = 11.0 \, \text{m}\) (final height)
- \(y_0 = 0.45 \, \text{m}\) (initial height)
- \(a = -9.8 \, \text{m/s}^2\) (acceleration due to gravity)

We can also use the vertical component of the final velocity to find time or initial velocity:
\[
v_{yf}^2 = v_{y0}^2 + 2a \Delta y
\]
where \(v_{yf} = 40.0 \, \text{m/s} \times \sin(12.9^\circ)\).

### Step 2: Horizontal motion analysis
For horizontal motion:
\[
v_x = \text{constant} = v_{fx} = v_f \times \cos(12.9^\circ)
\]
where \(v_f = 40.0 \, \text{m/s}\).

Since there is no acceleration in the horizontal direction, the horizontal distance traveled \(d_x = 103 \, \text{m}\) will help determine the time of flight and initial velocity.

---

I'll calculate the values for you next.### Results:
1. The **magnitude of the initial velocity** of the fragment is approximately **42.51 m/s**.
2. The **angle that the initial velocity makes with the horizontal** is approximately **23.47°**.

Would you like further explanation or more details on the steps? Here are some related questions:

1. How does the vertical displacement influence the initial velocity?
2. Why is horizontal velocity considered constant in projectile motion?
3. What other factors might affect the trajectory if air resistance were considered?
4. How would the problem change if the fragment hit obstacles before embedding in the tree?
5. Can this method be used to solve similar problems involving different angles of impact?

**Tip:** When solving projectile problems, separating horizontal and vertical motions simplifies calculations as horizontal motion is unaffected by gravity.

A 2.0 kg metal fragment recovered 103 m North of the scene of an explosion is found embedded in a tree 11.0 m above ground level. The metal fragment originates from an exploded propane tank and hit the tree without hitting anything else on the way. The fragment originated from 0.45 m above ground level. The damage on the tree indicates that the fragment struck at an angle 12.9° below the horizontal (as it was on its way DOWN as it made contact with the tree) with a speed of 40.0 m/s. What is the magnitude of the initial velocity of the fragment? What angle does the initial velocity make with respect to the horizontal?

Solve a projectile motion problem involving a metal fragment from an explosion. The fragment traveled 103 m horizontally and hit a tree 11.0 m above ground level. Learn how to calculate the initial velocity and angle using kinematic equations. Suitable for Grades 11-12 physics students.

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