Question
A rubber ball dropped on a hard surface takes a sequence of bounces, each one LaTeX: \frac{3}{5} as high as the preceding one. If this ball is dropped from a height of 10 feet, what is the total vertical distance it has traveled after it hits the
Answers
GPT-4o mini
To find the total vertical distance traveled by a rubber ball dropped from a height of 10 feet, where each bounce reaches \(\frac{3}{5}\) of the height of the previous bounce, we'll follow these steps:
1. **Initial Drop:**
The ball is dropped from an initial height of 10 feet.
2. **First Bounce:**
After the first drop, the ball bounces back to a height of:
\[
h_1 = 10 \times \frac{3}{5} = 6 \text{ feet}
\]
3. **Second Bounce:**
Then, it bounces again, reaching a height of:
\[
h_2 = 6 \times \frac{3}{5} = 3.6 \text{ feet}
\]
4. **General Pattern:**
The height after each bounce follows the formula:
\[
h_n = 10 \left( \frac{3}{5} \right)^n
\]
where \(n\) is the bounce number (starting from 0 for the initial drop).
5. **Vertical Distance Calculation:**
The total vertical distance \(D\) traveled can be calculated as follows:
- The distance for the initial drop: \(10\) feet down.
- The distance for the first bounce up and down: \(6\) feet up and \(6\) feet down.
- The distance for the second bounce up and down: \(3.6\) feet up and \(3.6\) feet down.
- This process continues for each subsequent bounce.
The total distance can be expressed as:
\[
D = 10 + 2 \times \sum_{n=0}^{\infty} h_n
\]
where the series represents the heights of the bounces.
6. **Sum of the Series:**
The infinite geometric series is given by:
\[
\sum_{n=0}^{\infty} h_n = \sum_{n=0}^{\infty} 10 \left( \frac{3}{5} \right)^n
\]
The first term \(a = 10\) and the common ratio \(r = \frac{3}{5}\). The sum of the infinite geometric series can be calculated using:
\[
S = \frac{a}{1 - r} = \frac{10}{1 - \frac{3}{5}} = \frac{10}{\frac{2}{5}} = 10 \times \frac{5}{2} = 25
\]
7. **Calculating the Total Distance:**
Now substituting back into the distance formula:
\[
D = 10 + 2 \times 25
\]
which evaluates to:
\[
D = 10 + 50 = 60 \text{ feet}
\]
Thus, the total vertical distance the ball has traveled after it hits the ground is \(\boxed{60}\) feet.
1. **Initial Drop:**
The ball is dropped from an initial height of 10 feet.
2. **First Bounce:**
After the first drop, the ball bounces back to a height of:
\[
h_1 = 10 \times \frac{3}{5} = 6 \text{ feet}
\]
3. **Second Bounce:**
Then, it bounces again, reaching a height of:
\[
h_2 = 6 \times \frac{3}{5} = 3.6 \text{ feet}
\]
4. **General Pattern:**
The height after each bounce follows the formula:
\[
h_n = 10 \left( \frac{3}{5} \right)^n
\]
where \(n\) is the bounce number (starting from 0 for the initial drop).
5. **Vertical Distance Calculation:**
The total vertical distance \(D\) traveled can be calculated as follows:
- The distance for the initial drop: \(10\) feet down.
- The distance for the first bounce up and down: \(6\) feet up and \(6\) feet down.
- The distance for the second bounce up and down: \(3.6\) feet up and \(3.6\) feet down.
- This process continues for each subsequent bounce.
The total distance can be expressed as:
\[
D = 10 + 2 \times \sum_{n=0}^{\infty} h_n
\]
where the series represents the heights of the bounces.
6. **Sum of the Series:**
The infinite geometric series is given by:
\[
\sum_{n=0}^{\infty} h_n = \sum_{n=0}^{\infty} 10 \left( \frac{3}{5} \right)^n
\]
The first term \(a = 10\) and the common ratio \(r = \frac{3}{5}\). The sum of the infinite geometric series can be calculated using:
\[
S = \frac{a}{1 - r} = \frac{10}{1 - \frac{3}{5}} = \frac{10}{\frac{2}{5}} = 10 \times \frac{5}{2} = 25
\]
7. **Calculating the Total Distance:**
Now substituting back into the distance formula:
\[
D = 10 + 2 \times 25
\]
which evaluates to:
\[
D = 10 + 50 = 60 \text{ feet}
\]
Thus, the total vertical distance the ball has traveled after it hits the ground is \(\boxed{60}\) feet.