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Part 1
A ball thrown straight up in the air travels 64 inches in the first tenth of a second.  In the next tenth of a​ second, the ball travels 60 inches.  After each additional tenth of a​ second, the ball travels 4 inches less than it did during the preceding tenth of a second.  How long will it take before the ball starts coming back​ down?  What is the total distance that the ball has traveled when it has reached its maximum​ height?
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Part 1
It will take 
  
enter your response here seconds before the ball starts coming back down.


Let's break down the problem systematically.

### Step 1: Understand the pattern of distances traveled
The ball's movement follows a pattern where:
- In the first \(0.1\) seconds, the ball travels \(64\) inches.
- In the next \(0.1\) seconds, the ball travels \(60\) inches.
- After that, for each additional \(0.1\) seconds, the distance decreases by \(4\) inches.

This forms an **arithmetic sequence** where:
- First term (\(a_1\)) = \(64\) inches,
- Common difference (\(d\)) = \(-4\) inches per \(0.1\) seconds.

### Step 2: Set the distance equal to 0 (or negative)
The ball will stop rising when the distance traveled in a \(0.1\)-second interval reaches zero or becomes negative. We can find when this happens using the formula for the \(n\)-th term of an arithmetic sequence:

\[
a_n = a_1 + (n-1) \cdot d
\]

Where:
- \(a_n\) is the distance traveled in the \(n\)-th \(0.1\) second,
- \(n\) is the number of time intervals,
- \(a_1 = 64\),
- \(d = -4\).

We need to find \(n\) when the distance \(a_n = 0\):

\[
0 = 64 + (n-1)(-4)
\]
\[
0 = 64 - 4(n-1)
\]
\[
4(n-1) = 64
\]
\[
n-1 = 16
\]
\[
n = 17
\]

### Step 3: Calculate the time before the ball starts coming down
Each \(n\) represents \(0.1\) seconds, so after \(n = 17\) intervals, the ball will stop rising.

Thus, the total time is:
\[
t = 17 \times 0.1 = 1.7 \, \text{seconds}
\]

### Step 4: Calculate the total distance traveled
The total distance is the sum of the distances traveled in each interval, which can be calculated by summing the first 17 terms of the arithmetic sequence.

The sum of the first \(n\) terms of an arithmetic sequence is given by:
\[
S_n = \frac{n}{2} \cdot (a_1 + a_n)
\]

Where:
- \(n = 17\),
- \(a_1 = 64\),
- \(a_n = 0\) (since it stops at 0).

Thus:
\[
S_{17} = \frac{17}{2} \cdot (64 + 0) = \frac{17}{2} \cdot 64 = 17 \cdot 32 = 544 \, \text{inches}
\]

### Final Answers:
- The ball will take **1.7 seconds** before it starts coming back down.
- The total distance traveled by the ball when it reaches its maximum height is **544 inches**.

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:
1. How would the solution change if the distance decrease per interval was different?
2. What if the ball traveled 68 inches in the first \(0.1\) seconds? How would this affect the results?
3. Can you derive a general formula for the time it takes for a ball to reach its maximum height given different initial velocities?
4. How would you calculate the total distance if air resistance was a factor?
5. What is the relationship between the distance pattern and the acceleration due to gravity?

**Tip:** The sum of an arithmetic sequence can quickly estimate the total distance in similar problems involving constant acceleration.

A ball thrown straight up in the air travels 64 inches in the first tenth of a second. In the next tenth of a​ second, the ball travels 60 inches. After each additional tenth of a​ second, the ball travels 4 inches less than it did during the preceding tenth of a second. How long will it take before the ball starts coming back​ down? What is the total distance that the ball has traveled when it has reached its maximum​ height?

Learn how to solve a problem involving a ball thrown upwards, with distances decreasing in an arithmetic sequence. This problem explores kinematics and arithmetic progressions, and includes a step-by-step solution to find the time when the ball starts coming down and the total distance traveled. Suitable for Grades 9-11.

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