A paper airplane is thrown from the top of a hill and travels horizontally at 9 feet per seconds. If the plane desends 1 foot for every 3 feet travelled horizontally, how may feet has the plane descended after 5 seconds of travel?

Options

A. `3`
B. `10`
C. `15`
D. `20`

Answer

To solve the problem step-by-step, we will follow these steps:

### Step 1: Calculate the horizontal distance traveled by the airplane.
The airplane travels horizontally at a speed of 9 feet per second for a duration of 5 seconds.

\[ \text{Distance} = \text{Speed} \times \text{Time} \]

Substituting the values:

\[ \text{Distance} = 9 \, \text{feet/second} \times 5 \, \text{seconds} = 45 \, \text{feet} \]

### Step 2: Determine the descent rate of the airplane.
The problem states that for every 3 feet traveled horizontally, the airplane descends 1 foot. We need to find out how much the airplane descends after traveling 45 feet horizontally.

### Step 3: Set up a proportion to find the descent.
Let \( x \) be the total descent in feet after traveling 45 feet. We can set up the proportion based on the given descent rate:

\[ \frac{1 \, \text{foot}}{3 \, \text{feet}} = \frac{x \, \text{feet}}{45 \, \text{feet}} \]

### Step 4: Cross-multiply to solve for \( x \).
Cross-multiplying gives us:

\[ 1 \cdot 45 = 3 \cdot x \]

This simplifies to:

\[ 45 = 3x \]

### Step 5: Solve for \( x \).
Now, divide both sides by 3:

\[ x = \frac{45}{3} = 15 \, \text{feet} \]

### Conclusion:
The airplane has descended **15 feet** after 5 seconds of travel.

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