The force of the wind on a sail varies jointly as the area of the sail and the square of the wind velocity. On a sail of area 50 square yards , the force of a 15-mile-per-hour wind is 45 pounds . Find the force on the sail if the wind increases to 45 miles per hour.

Options

A. 135 pounds
B. 225 pounds
C. 405 pounds
D. 450 pounds

Explanation

To solve the problem, we need to find the force on the sail when the wind velocity increases to 45 miles per hour, given that the force of the wind on a sail varies jointly as the area of the sail and the square of the wind velocity.

### Step-by-Step Solution:

1. **Understand the relationship**:
The force \( F \) on the sail varies jointly as the area \( A \) of the sail and the square of the wind velocity \( v \). This can be expressed mathematically as:
\[ F = k \cdot A \cdot v^2 \]
where \( k \) is a constant of proportionality.

2. **Identify known values**:
From the problem, we have:
- Area \( A = 50 \) square yards
- Initial wind velocity \( v_1 = 15 \) miles per hour
- Initial force \( F_1 = 45 \) pounds

3. **Find the constant \( k \)**:
Using the known values, we can substitute into the equation to find \( k \):
\[ 45 = k \cdot 50 \cdot (15)^2 \]
\[ 45 = k \cdot 50 \cdot 225 \]
\[ 45 = k \cdot 11250 \]
\[ k = \frac{45}{11250} = \frac{1}{250} \]

4. **Calculate the new force \( F_2 \)**:
Now, we want to find the force when the wind velocity increases to \( v_2 = 45 \) miles per hour. We can use the formula again:
\[ F_2 = k \cdot A \cdot v_2^2 \]
Substituting the values we have:
\[ F_2 = \frac{1}{250} \cdot 50 \cdot (45)^2 \]
\[ F_2 = \frac{1}{250} \cdot 50 \cdot 2025 \]
\[ F_2 = \frac{1}{250} \cdot 101250 \]
\[ F_2 = 405 \]

5. **Conclusion**:
The force on the sail when the wind velocity increases to 45 miles per hour is **405 pounds**.