Here is the question:

a ball is dropped from a height of 3m and bounces on the ground . At the top of each bounce, the ball eaches 60% of its previous height. Calculate the total distance travelled by the ball when it hits the ground for the fifth time?

I know you need the n, r and a values.
I got r=0.6 n=5 and a=3 but every where i look they have a=6 and subtract 3 off the final answer? why?

because there was no bounce at first - just a drop.

When it hits the ground for the 5th time, it has made a single 3-foot drop, and then 4 round-trip bounces.

To calculate the total distance traveled by the ball when it hits the ground for the fifth time, we need to find the sum of the distances covered during each bounce.

Given:
Initial height (a) = 3m
Bounce ratio (r) = 0.6
Number of bounces (n) = 5

To clarify, the given values are indeed, a = 3, r = 0.6, and n = 5.

To calculate the distance covered during each bounce, we can use the formula for the sum of a geometric series:

S = a * (1 - r^n) / (1 - r)

Let's plug in the values:

a = 3
r = 0.6
n = 5

S = 3 * (1 - 0.6^5) / (1 - 0.6)

Simplifying this expression:

S = 3 * (1 - 0.07776) / (1 - 0.6)
= 3 * 0.92224 / 0.4
= 6.96672 / 0.4
= 17.4168

Therefore, the total distance traveled by the ball when it hits the ground for the fifth time is approximately 17.42 meters.

It seems there may be a confusion regarding the values given elsewhere. However, based on the information you provided, the value for a is indeed 3, and you do not need to subtract 3 from the final answer.

To solve this problem, we need to find the total distance traveled by the ball when it hits the ground for the fifth time. Let's break down the process step by step:

1. Identify the pattern: The problem states that at the top of each bounce, the ball reaches 60% of its previous height. This means that the ball's height decreases by 40% each time it bounces back up.

2. Calculate the ball's height at each bounce:
- On the first bounce, the ball reaches 60% of its initial height (3m). So, the height after the first bounce is 0.6 * 3 = 1.8m.
- On the second bounce, the ball reaches 60% of its height after the first bounce (1.8m). So, the height after the second bounce is 0.6 * 1.8 = 1.08m.
- Continue this pattern for each bounce until the fifth one.

3. Find the total distance traveled: The total distance traveled by the ball is the sum of all the up and down movements, including the initial drop. Each up and down movement covers twice the height change.

Let's calculate the total distance:

- The initial drop is 3m.
- The first up and down movement is 2 * (3m - 1.8m) = 2 * 1.2m = 2.4m.
- The second up and down movement is 2 * (1.8m - 1.08m) = 2 * 0.72m = 1.44m.
- Continue this calculation for each up and down movement until the fifth one.

Adding up these distances for the five up and down movements, we get:
3m + 2.4m + 1.44m + ... (keep going for the remaining up and down movements)

Now, the tricky part is to find the formula for the sum of this geometric series. The formula is:

Sum = a * (1 - r^n) / (1 - r)

In this formula:
- a represents the first term of the series, which is the height of the initial drop (3m).
- r represents the common ratio between the terms of the series (0.6).
- n represents the number of terms in the series (5).

Plugging in the values, we have:
Sum = 3 * (1 - 0.6^5) / (1 - 0.6)

Now, let's calculate the sum. Substitute the values and perform the calculations:

Sum = 3 * (1 - 0.07776) / (0.4)
= 3 * 0.92224 / 0.4
= 6.96768 / 0.4
= 17.4192 meters (rounded to four decimal places)

So, the total distance traveled by the ball when it hits the ground for the fifth time is approximately 17.4192 meters.