A bird flies overhead from where you stand at an altitude of 350.0ĵ m and at a velocity horizontal to the ground of 30.0î m/s. The bird has a mass of 2.1 kg. The radius vector to the bird makes an angle 𝜃 with respect to the ground. The radius vector to the bird lies in the xy-plane. What is the bird's angular momentum (in kg · m2/s) about the point where you are standing? (Express your answer in vector form.)
 l  =    kg · m2/s

To find the angular momentum (\( \mathbf{l} \)) of the bird about the given point, we use the formula:

\[
\mathbf{l} = \mathbf{r} \times \mathbf{p}
\]

Where:
- \( \mathbf{r} \) is the position vector of the bird relative to the point,
- \( \mathbf{p} \) is the linear momentum of the bird, and
- \( \times \) denotes the cross product.

### Given Data:
1. Altitude of the bird: \( z = 350.0 \, \mathrm{m} \).
2. Horizontal velocity of the bird: \( \mathbf{v} = 30.0 \hat{i} \, \mathrm{m/s} \).
3. Mass of the bird: \( m = 2.1 \, \mathrm{kg} \).
4. \( \mathbf{r} \) lies in the xy-plane and makes an angle \( \theta \) with the x-axis.

#### Step 1: Define \( \mathbf{r} \)
The radius vector \( \mathbf{r} \) is given as:
\[
\mathbf{r} = x \hat{i} + z \hat{j}
\]
Here:
- \( x = r \cos\theta \),
- \( z = r \sin\theta \), where \( z = 350.0 \, \mathrm{m} \).

From the geometry of the situation, \( x = \sqrt{r^2 - z^2} \), so:
\[
x = \sqrt{r^2 - 350^2}
\]

#### Step 2: Define \( \mathbf{p} \)
The linear momentum of the bird is:
\[
\mathbf{p} = m \mathbf{v} = (2.1)(30.0 \hat{i}) = 63.0 \hat{i} \, \mathrm{kg \cdot m/s}.
\]

#### Step 3: Calculate \( \mathbf{l} = \mathbf{r} \times \mathbf{p} \)
The cross product is calculated as:
\[
\mathbf{l} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
x & 0 & z \\
63.0 & 0 & 0 \\
\end{vmatrix}
\]

This expands to:
\[
\mathbf{l} = \hat{i} \begin{vmatrix}
0 & z \\
0 & 0
\end{vmatrix}
- \hat{j} \begin{vmatrix}
x & z \\
63.0 & 0
\end{vmatrix}
+ \hat{k} \begin{vmatrix}
x & 0 \\
63.0 & 0
\end{vmatrix}
\]

\[
\mathbf{l} = 0 \hat{i} - \hat{j} \left(0 - 63.0z\right) + \hat{k} \left(0 - 63.0(0)\right)
\]

\[
\mathbf{l} = 63.0z \hat{j}
\]

Substitute \( z = 350.0 \, \mathrm{m} \):
\[
\mathbf{l} = 63.0(350.0) \hat{j} = 22050.0 \hat{j} \, \mathrm{kg \cdot m^2/s}.
\]

### Final Answer:
\[
\mathbf{l} = 22050.0 \hat{j} \, \mathrm{kg \cdot m^2/s}.
\]

Let me know if you have further questions or need clarifications! 😊

---

1. What is the significance of angular momentum in rotational motion?
2. How does the cross product influence the direction of angular momentum?
3. What happens to the angular momentum if the bird's velocity changes direction?
4. Can angular momentum be conserved in this scenario? Why or why not?
5. How would the result change if the radius vector extended into the z-axis?

**Tip:** Always verify the dimensions of your quantities in physics problems to ensure consistency in units!

A bird flies overhead from where you stand at an altitude of 350.0ĵ m and at a velocity horizontal to the ground of 30.0î m/s. The bird has a mass of 2.1 kg. The radius vector to the bird makes an angle 𝜃 with respect to the ground. The radius vector to the bird lies in the xy-plane. What is the bird's angular momentum (in kg · m^2/s) about the point where you are standing? (Express your answer in vector form.)

Discover how to calculate the angular momentum of a bird flying at 30.0 m/s overhead at an altitude of 350.0 m. This detailed solution uses vector cross products, linear momentum, and physics kinematics, ideal for Grades 11-12 or college-level physics.

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