At a particular instant, a stationary observer on the ground sees a package falling with a speed `v_(1)` at an angle `theta` to the vertical. To a pilot flying horizontally with a constant speed v relative to the ground, the package appears to be falling verically with a speed `v_(2)` (at that same instant).
The speed of the pilot relative to the ground (v) is

Options

A. `(V_(1)^(2)+v_(2)^(2))^(1//2)`
B. `(v_(1)-v_(2))(v_(2)-v_(1))`
C. `(v_(1)^(2)-v_(2)^(2))^(1//2)`
D. `v_(1)-v_(2)`

Answer

To solve the problem, we need to analyze the motion of the package as observed by both the stationary observer on the ground and the pilot flying horizontally.

### Step-by-Step Solution:

1. **Understanding the Situation**:
- A package is falling with a speed \( v_1 \) at an angle \( \theta \) to the vertical as observed by a stationary observer on the ground.
- The pilot flying horizontally with speed \( v \) sees the package falling vertically with speed \( v_2 \).

2. **Setting Up the Vectors**:
- The velocity of the package can be broken down into two components:
- Vertical component: \( v_{1y} = v_1 \cos(\theta) \)
- Horizontal component: \( v_{1x} = v_1 \sin(\theta) \)

3. **Relative Motion**:
- For the pilot, the package appears to fall vertically. This means that the horizontal component of the package's velocity must be equal to the horizontal speed of the pilot, \( v \).
- Therefore, we can equate the horizontal components:
\[ v_{1x} = v \]
- This gives us:
\[ v = v_1 \sin(\theta) \]

4. **Vertical Component**:
- The vertical component of the package's velocity as observed by the pilot is \( v_{1y} - v \) (since the pilot is moving horizontally).
- Since the package appears to fall vertically with speed \( v_2 \), we can write:
\[ v_{1y} = v_2 \]
- Therefore, we have:
\[ v_1 \cos(\theta) = v_2 \]

5. **Finding the Speed of the Pilot**:
- We now have two equations:
1. \( v = v_1 \sin(\theta) \)
2. \( v_2 = v_1 \cos(\theta) \)
- We can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = \frac{v_2}{\cos(\theta)} \]
- Substituting this into the equation for \( v \):
\[ v = \left(\frac{v_2}{\cos(\theta)}\right) \sin(\theta) \]
- Simplifying gives:
\[ v = v_2 \tan(\theta) \]

6. **Final Expression**:
- The speed of the pilot relative to the ground is:
\[ v = \sqrt{v_1^2 - v_2^2} \]

### Final Answer:
The speed of the pilot relative to the ground \( v \) is given by:
\[ v = \sqrt{v_1^2 - v_2^2} \]