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 need to analyze the motion of the mouse after it is released by the eagle. We will use the concepts of relative velocity and vector addition to find the speed of the eagle with respect to the observer on the ground.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The eagle is flying horizontally with a constant velocity \( V_0 \).
- When the eagle releases the mouse, the mouse has the same horizontal velocity \( V_0 \) as the eagle at the moment of release.

2. **Velocity of the Mouse**:
- From the eagle's perspective, the mouse falls vertically downwards with a speed \( V_1 \). This means that the vertical component of the mouse's velocity is \( V_1 \) and its horizontal component remains \( V_0 \).

3. **Observer's Perspective**:
- From the observer on the ground, the mouse will have both horizontal and vertical components of velocity. The horizontal component is \( V_0 \) and the vertical component is \( V_y = g t \), where \( g \) is the acceleration due to gravity and \( t \) is the time since the mouse was released.

4. **Resultant Velocity**:
- The mouse's velocity as seen by the observer can be represented as a vector with horizontal component \( V_0 \) and vertical component \( V_1 \). The resultant velocity \( V_2 \) can be found using the Pythagorean theorem:
\[ V_2 = \sqrt{V_0^2 + V_1^2} \]

5. **Finding the Speed of the Eagle**:
- We know that the horizontal component of the mouse's velocity as seen by the observer is \( V_0 \) and the vertical component is \( V_1 \). Therefore, we can express the relationship between these velocities:
\[ V_0 = \sqrt{V_2^2 - V_1^2} \]
- This equation gives us the speed of the eagle with respect to the observer on the ground.

### Final Answer:
The speed of the eagle with respect to the observer on the ground is given by:
\[ V_0 = \sqrt{V_2^2 - V_1^2} \]