To determine the vertical distance traveled by the ball when it hits the ground for the fourth time, let's carefully analyze the process step by step.
1. Initial Drop:
The ball is dropped from an initial height of 5 feet. This is the first distance traveled by the ball.
[tex]\[
\text{Initial Drop} = 5 \text{ ft}
\][/tex]
2. First Bounce:
After the initial drop, the ball bounces to 60% of its initial height, which is:
[tex]\[
0.6 \times 5 \text{ ft} = 3 \text{ ft}
\][/tex]
The ball travels this same distance (3 ft) again when it falls back to the ground, making the total distance for the first bounce:
[tex]\[
2 \times 3 \text{ ft} = 6 \text{ ft}
\][/tex]
3. Second Bounce:
The ball now bounces to 60% of 3 feet:
[tex]\[
0.6 \times 3 \text{ ft} = 1.8 \text{ ft}
\][/tex]
The ball travels this distance twice again (up and down), making the total distance for the second bounce:
[tex]\[
2 \times 1.8 \text{ ft} = 3.6 \text{ ft}
\][/tex]
4. Third Bounce:
The ball now bounces to 60% of 1.8 feet:
[tex]\[
0.6 \times 1.8 \text{ ft} = 1.08 \text{ ft}
\][/tex]
The ball travels this distance twice (up and down), making the total distance for the third bounce:
[tex]\[
2 \times 1.08 \text{ ft} = 2.16 \text{ ft}
\][/tex]
5. Fourth Bounce:
The ball now bounces to 60% of 1.08 feet:
[tex]\[
0.6 \times 1.08 \text{ ft} = 0.648 \text{ ft}
\][/tex]
The ball travels this distance just once when it falls back to the ground, making the total distance for the fourth bounce:
[tex]\[
0.648 \text{ ft}
\][/tex]
Now, we sum up all these distances to find the total vertical distance traveled by the ball when it hits the ground for the fourth time:
[tex]\[
\begin{align*}
\text{Total Distance} &= \text{Initial Drop} + \sum_{n=1}^4 \text{Subsequent Bounces} \\
&= 5 + 2 \times 3 + 2 \times 1.8 + 2 \times 1.08 + 0.648 \\
&= 5 + 6 + 3.6 + 2.16 + 0.648 \\
&= 5 + 11.408 + 3.6 + 0.648 \\
&= 11.408 + 3.6 + 2.808 \\
&= 18.055999999999997 \text{ ft}
\end{align*}
\][/tex]
Therefore, the correct approach that results in 18.06 ft is when one uses:
[tex]\[
5 + \sum_{n=1}^4 2 \times (0.6)^n \times 5
\][/tex]
Thus, the correct statement is:
"One should use [tex]\(5 + \sum_{n=1}^4 2(0.6)^n(5)\)[/tex], which results in 18.06 ft for the vertical distance traveled."
