A ball is dropped from a height of 15 inches.  On each bounce, it rises to 60% of its previous height.  

When it hits the ground for the eighth time, how far has it traveled? 

To solve this problem, we need to calculate the total distance traveled by the ball after it hits the ground for the eighth time. The ball is dropped from a height of 15 inches and, on each bounce, it rises to 60% of its previous height. 

### Step-by-Step Solution

1. **Understanding the motion of the ball:**

   The ball falls from a height of 15 inches initially. After hitting the ground, it bounces back up to 60% of its previous height. It then falls back down from this height, and this pattern continues.

2. **Distance Calculation:**

   The distance traveled by the ball can be separated into two parts:
   - The initial fall from the height of 15 inches.
   - The subsequent bounces up and down.

3. **Breaking down the distance:**

   - The initial drop is 15 inches.
   - The first bounce rises to \(15 \times 0.6 = 9\) inches and falls back down the same distance (9 inches).
   - The second bounce rises to \(9 \times 0.6 = 5.4\) inches and falls back down the same distance (5.4 inches).
   - This pattern continues.

   The height after the \(n\)-th bounce is given by:
   \[
   h_n = 15 \times (0.6)^n
   \]

4. **Total distance traveled after 8 bounces:**

   The total distance \(D\) traveled is the sum of all the distances traveled up to the 8th bounce. This includes:

   - The initial drop of 15 inches.
   - The distance of the ball going up and down for each bounce.

   Therefore, the total distance traveled can be expressed as:
   \[
   D = 15 + 2 \times (9 + 5.4 + 3.24 + 1.944 + \ldots + h_7)
   \]

   The distances form a geometric series for the bounces, where each term represents the height to which the ball rises after each bounce.

5. **Calculate the total distance using the geometric series formula:**

   The series for the height of each bounce (after the initial drop) is:
   \[
   S = 9 + 5.4 + 3.24 + 1.944 + \ldots + h_7
   \]

   This is a geometric series with the first term \(a = 9\) and a common ratio \(r = 0.6\).

   The sum of the first \(n\) terms of a geometric series is given by:
   \[
   S_n = a \frac{1 - r^n}{1 - r}
   \]

   For \(n = 7\), \(a = 9\), and \(r = 0.6\), we find:

   \[
   S_7 = 9 \frac{1 - (0.6)^7}{1 - 0.6}
   \]

   Now, we can compute this.The sum of the distances for the bounces up to the 7th bounce is approximately \( S_7 = 21.87 \) inches.

6. **Total distance traveled by the ball:**

   Now, the total distance \( D \) the ball has traveled when it hits the ground for the 8th time is:

   \[
   D = 15 + 2 \times S_7
   \]

   Substituting the value of \( S_7 \):

   \[
   D = 15 + 2 \times 21.87 = 15 + 43.74 = 58.74 \text{ inches}
   \]

### Conclusion

The total distance traveled by the ball when it hits the ground for the eighth time is **58.74 inches**.

Would you like further details on this solution, or do you have any questions?

### Related Questions to Explore:

1. How would the total distance change if the bounce height were 70% instead of 60%?
2. What would be the distance traveled after the ball hits the ground for the 10th time?
3. Can you find the total distance if the initial height was 20 inches instead of 15 inches?
4. How would the total distance change if the ball were dropped from a different height?
5. What is the significance of the common ratio in determining the total distance traveled?

**Tip:** Geometric series formulas are useful for problems involving repeated multiplication, like depreciation, population growth, or bouncing balls!

A ball is dropped from a height of 15 inches. On each bounce, it rises to 60% of its previous height. When it hits the ground for the eighth time, how far has it traveled?

Find out how to calculate the total distance traveled by a ball dropped from a height of 15 inches and rising to 60% of its previous height on each bounce. Detailed step-by-step solution, suitable for students in Grades 9-11, covering geometric series and percentage concepts.

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