A stone is dropped from the top of a building in a high wind. The wind exerts a steady horizontal force on the stone as it falls. The path of the string is

Options

A. a straight line
B. a parabola
C. a circle
D. another complicated path

Explanation

To solve the problem of a stone dropped from the top of a building in a high wind, we need to analyze the motion of the stone under the influence of gravity and the horizontal force exerted by the wind. Here’s a step-by-step solution:

### Step 1: Analyze the Forces Acting on the Stone
When the stone is dropped, two forces act on it:
- The gravitational force acting downwards (weight of the stone, \( mg \)).
- The horizontal force due to the wind, which we can denote as \( F \).

### Step 2: Determine the Accelerations
The stone experiences two components of acceleration:
- Vertical acceleration (\( a_y \)): This is due to gravity and is equal to \( g \) (approximately \( 9.81 \, \text{m/s}^2 \)).
- Horizontal acceleration (\( a_x \)): This is due to the wind's force. According to Newton's second law, \( a_x = \frac{F}{m} \).

### Step 3: Write the Equations of Motion
Using the equations of motion, we can express the vertical and horizontal displacements:
- For vertical motion (y-direction):
\[ y = \frac{1}{2} g t^2 \]
- For horizontal motion (x-direction):
\[ x = \frac{1}{2} a_x t^2 = \frac{1}{2} \left(\frac{F}{m}\right) t^2 \]

### Step 4: Relate the Two Displacements
Now, we can relate \( y \) and \( x \):
\[ \frac{y}{x} = \frac{\frac{1}{2} g t^2}{\frac{1}{2} \left(\frac{F}{m}\right) t^2} \]
This simplifies to:
\[ \frac{y}{x} = \frac{g}{\frac{F}{m}} = \frac{mg}{F} \]
This indicates that \( y \) is proportional to \( x \).

### Step 5: Conclusion on the Path of the Stone
Since \( y \) is proportional to \( x \), the path of the stone will be a straight line. Therefore, the trajectory of the stone will be a straight line inclined at an angle to the vertical.

### Final Answer
The path of the stone is a straight line.

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