1. A ball is dropped from a height of 18m, and each rebound is 1/3 of the previous distance. a. How high is the ball in its 4th rebound? b. What is the total distance the ball has travelled if you catch it after reaches the top of the 5th rebound?

### Part (a): How high is the ball in its 4th rebound?

The ball is dropped from a height of 18 meters, and each rebound is \( \frac{1}{3} \) of the previous height. To find the height of the ball on the 4th rebound, we use a geometric sequence, where the initial height \( h_0 = 18 \) meters and the common ratio \( r = \frac{1}{3} \).

The height of the ball after the \( n \)-th rebound is given by:
\[
h_n = h_0 \times r^n
\]

For the 4th rebound \( n = 4 \):
\[
h_4 = 18 \times \left( \frac{1}{3} \right)^4 = 18 \times \frac{1}{81} = \frac{18}{81} = 0.222 \text{ meters}
\]

So, the height of the ball on its 4th rebound is **0.222 meters**.

### Part (b): What is the total distance traveled after the 5th rebound?

To find the total distance the ball has traveled when it reaches the top of the 5th rebound, we need to account for both the distance the ball falls and the distance it rebounds.

Each time the ball falls from a height, it then rebounds to a fraction of that height. Therefore, the total distance includes:
- The initial drop from 18 meters.
- The series of upward and downward travels from each rebound.

#### Step 1: Distance for the initial fall and first rebound:
- Initial drop: 18 meters
- First rebound: \( 18 \times \frac{1}{3} = 6 \) meters

#### Step 2: Geometric series for subsequent distances:
For every subsequent rebound after the first, the ball falls the same height it rebounds, and the height of each rebound follows a geometric sequence. The total distance for these rebounds includes both the upward and downward movements.

We need to sum the distances for the first 5 rebounds.

\[
\text{Distance after first fall and rebound} = 18 + 6 = 24 \text{ meters}
\]
For the subsequent 4 rebounds:
The total distance for the upward and downward movement is:
\[
\text{Total for n rebounds} = 2 \times \left(6 + 6 \times \frac{1}{3} + 6 \times \left(\frac{1}{3}\right)^2 + 6 \times \left(\frac{1}{3}\right)^3 + 6 \times \left(\frac{1}{3}\right)^4\right)
\]
Which simplifies to:
\[
2 \times (6 + 2 + 0.6667 + 0.2222 + 0.07407) = 2 \times 8.963
\]
\[
= 17.926 \text{ meters}
\]

#### Step 3: Adding the initial drop:
\[
\text{Total distance} = 24 + 17.926 = 41.926 \text{ meters}
\]

Thus, the total distance the ball has traveled after being caught at the top of the 5th rebound is approximately **41.93 meters**.

---

Would you like further details or have any questions?

Here are some related questions:
1. What would be the height of the ball after 6th rebound?
2. How would the total distance change if the rebound ratio was \( \frac{1}{2} \) instead of \( \frac{1}{3} \)?
3. How does changing the initial height affect the total distance covered after 5 rebounds?
4. Can you derive a general formula for the total distance after \( n \) rebounds?
5. How would you model this situation graphically using a geometric series?

**Tip:** When dealing with repetitive rebounds or cycles, geometric series provide a handy way to sum up multiple terms efficiently.

A ball is dropped from a height of 18m, and each rebound is 1/3 of the previous distance. a. How high is the ball in its 4th rebound? b. What is the total distance the ball has travelled if you catch it after it reaches the top of the 5th rebound?

Solve a geometric sequence problem involving a ball dropped from 18m with each rebound being 1/3 of the previous height. Learn how to calculate the height of the ball after the 4th rebound and the total distance traveled after the 5th rebound. Suitable for high school students learning geometric series.

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