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 from both the ground and the pilot's perspective. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Situation
We have a package falling with a speed \( v_1 \) at an angle \( \theta \) to the vertical. The pilot is flying horizontally with a speed \( v \) relative to the ground. To the pilot, the package appears to be falling vertically with a speed \( v_2 \).

### Step 2: Set Up the Coordinate System
Let's define our coordinate system:
- The vertical direction (downward) is negative \( y \).
- The horizontal direction (to the right) is positive \( x \).

### Step 3: Express the Velocity of the Package
The velocity of the package with respect to the ground can be broken down into its components:
- The horizontal component (along \( x \)): \( v_{1x} = v_1 \sin \theta \) (to the left, hence negative)
- The vertical component (along \( y \)): \( v_{1y} = -v_1 \cos \theta \) (downward)

Thus, the velocity of the package with respect to the ground is:
\[ \vec{v}_{pg} = -v_1 \sin \theta \hat{i} - v_1 \cos \theta \hat{j} \]

### Step 4: Express the Velocity of the Pilot
The velocity of the pilot with respect to the ground is:
\[ \vec{v}_{p} = v \hat{i} \]

### Step 5: Use the Relative Velocity Equation
According to the relative motion formula:
\[ \vec{v}_{pg} - \vec{v}_{p} = \vec{v}_{p\text{g}} \]
Where \( \vec{v}_{p\text{g}} \) is the velocity of the package as seen by the pilot, which is purely vertical and equals \( -v_2 \hat{j} \).

Substituting the expressions we have:
\[ (-v_1 \sin \theta \hat{i} - v_1 \cos \theta \hat{j}) - (v \hat{i}) = -v_2 \hat{j} \]

### Step 6: Equate Components
Now, we can equate the components of the vectors:
1. For the \( \hat{i} \) (horizontal) component:
\[ -v_1 \sin \theta - v = 0 \quad \Rightarrow \quad v = -v_1 \sin \theta \quad \text{(1)} \]
2. For the \( \hat{j} \) (vertical) component:
\[ -v_1 \cos \theta = -v_2 \quad \Rightarrow \quad v_2 = v_1 \cos \theta \quad \text{(2)} \]

### Step 7: Find the Relationship Between \( v_1 \), \( v_2 \), and \( \theta \)
From equation (2), we can express \( \cos \theta \):
\[ \cos \theta = \frac{v_2}{v_1} \]

Using the Pythagorean identity, we can find \( \sin \theta \):
\[ \sin^2 \theta + \cos^2 \theta = 1 \quad \Rightarrow \quad \sin^2 \theta = 1 - \left(\frac{v_2}{v_1}\right)^2 \]
Thus,
\[ \sin \theta = \sqrt{1 - \frac{v_2^2}{v_1^2}} = \frac{\sqrt{v_1^2 - v_2^2}}{v_1} \]

### Step 8: Substitute Back to Find \( v \)
Substituting \( \sin \theta \) back into equation (1):
\[ v = -v_1 \sin \theta = -v_1 \left(\frac{\sqrt{v_1^2 - v_2^2}}{v_1}\right) = -\sqrt{v_1^2 - v_2^2} \]

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