An eagle flies at constant velocity horizontally across the sky, carrying a mouse and releases the mouse while in flight . From the eagle's perspective , the mouse falls vertically with speed `v_1` From an observer on the ground's perspective , the mouse falls at an angle with speed `v_2` what is the speed of the eagle with respect to the observer on the ground ?

Options

A. `v_1+v_2`
B. `v_1-v_2`
C. `sqrt(v_1^2-v_2^2)`
D. `sqrt(v_2^2-v_1^2)`

Answer

To solve the problem, we will analyze the situation using vector components and the Pythagorean theorem.

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- An eagle is flying horizontally with a constant velocity \( V_E \) and releases a mouse.
- From the eagle's perspective, the mouse falls vertically downwards with speed \( v_1 \).
- From the observer on the ground's perspective, the mouse falls at an angle with speed \( v_2 \).

2. **Setting Up the Coordinate System**:
- Let's assume the eagle is flying in the positive x-direction (horizontal).
- The mouse, when released, has a vertical velocity \( v_1 \) (downward) and a horizontal velocity \( V_E \) (same as the eagle's horizontal speed).

3. **Analyzing the Mouse's Motion**:
- From the ground observer's perspective, the mouse has two components of velocity:
- Horizontal component: \( V_E \)
- Vertical component: \( v_1 \)

4. **Using the Resultant Velocity**:
- The resultant velocity \( v_2 \) of the mouse from the ground observer's perspective can be calculated using the Pythagorean theorem:
\[ v_2 = \sqrt{V_E^2 + v_1^2} \]

5. **Rearranging the Equation**:
- To find the speed of the eagle \( V_E \) in terms of \( v_1 \) and \( v_2 \), we can rearrange the equation:
\[ V_E^2 = v_2^2 - v_1^2 \]
- Taking the square root gives:
\[ V_E = \sqrt{v_2^2 - v_1^2} \]

6. **Final Expression**:
- Thus, the speed of the eagle with respect to the observer on the ground is:
\[ V_E = \sqrt{v_2^2 - v_1^2} \]