When a certain rubber ball is dropped, it rebounds 75 percent as high as the distance it fell. What is...

1. Translate the problem requirements: The ball drops from height h, bounces to \(75\%\) of h, falls again, bounces to \(75\%\) of that height, then falls a final time. We need to find the original height h given that total vertical distance traveled is 2,146 cm.
2. Map out the ball's complete journey: Identify each segment of the ball's path - the initial drop, first bounce up and down, second bounce up and down - to understand what distances contribute to the total.
3. Express each distance in terms of the original height: Convert each segment of the journey into an expression involving the unknown original height h.
4. Set up and solve the total distance equation: Add all distance segments to equal 2,146 cm and solve for h, using approximation as needed given the answer choices are well-spaced.

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what's happening step by step. We have a rubber ball that gets dropped from some height - let's call this original height 'h'. When it hits the ground, it doesn't just stay there - it bounces back up, but only to \(75\%\) of the height it just fell from.

The key information we have is:

• The ball rebounds to \(75\%\) of the height it fell from each time
• By the time it hits the ground for the third time, it has traveled 2,146 cm total vertically
• We need to find the original drop height 'h'

Think of it like this: the ball goes down, bounces up (but not as high), comes back down, bounces up again (even lower), then comes down one final time to hit the ground for the third time.

Process Skill: TRANSLATE - Converting the bounce behavior and total distance into a clear mathematical setup

2. Map out the ball's complete journey

Let's trace the ball's path segment by segment. Imagine you're watching this ball in slow motion:

Journey breakdown:

• First drop: Ball falls from height h down to the ground (distance = h)
• First bounce: Ball bounces back up to \(75\%\) of h, which is \(0.75\mathrm{h}\) (distance = \(0.75\mathrm{h}\))
• Second drop: Ball falls back down from \(0.75\mathrm{h}\) to the ground (distance = \(0.75\mathrm{h}\))
• Second bounce: Ball bounces up to \(75\%\) of \(0.75\mathrm{h}\), which is \(0.75 \times 0.75\mathrm{h} = 0.5625\mathrm{h}\) (distance = \(0.5625\mathrm{h}\))
• Final drop: Ball falls from \(0.5625\mathrm{h}\) to the ground for the third time (distance = \(0.5625\mathrm{h}\))

Notice how each bounce gets progressively smaller - this makes physical sense because the ball loses energy with each bounce.

3. Express each distance in terms of the original height

Now let's add up all these individual distances to get our total:

Total distance traveled = All the individual segments

• First drop: h
• First bounce up: \(0.75\mathrm{h}\)
• Second drop down: \(0.75\mathrm{h}\)
• Second bounce up: \(0.75 \times 0.75\mathrm{h} = 0.5625\mathrm{h}\)
• Final drop down: \(0.5625\mathrm{h}\)

Total = \(\mathrm{h} + 0.75\mathrm{h} + 0.75\mathrm{h} + 0.5625\mathrm{h} + 0.5625\mathrm{h}\)

Let's group the terms to make this easier:
Total = \(\mathrm{h} + 2(0.75\mathrm{h}) + 2(0.5625\mathrm{h})\)
Total = \(\mathrm{h} + 1.5\mathrm{h} + 1.125\mathrm{h}\)
Total = \(3.625\mathrm{h}\)

Process Skill: VISUALIZE - Breaking down the complex motion into manageable segments that can be easily tracked

4. Set up and solve the total distance equation

We know the total distance traveled is 2,146 cm, so:

\(3.625\mathrm{h} = 2146\)

To solve for h:
\(\mathrm{h} = 2146 \div 3.625\)
\(\mathrm{h} = 592.1...\) cm

Since we need to choose from the given answer choices and the problem asks for an approximate height, let's see which choice is closest:

1. 400 cm (too low)
2. 500 cm (too low)
3. 600 cm (very close to our 592 cm!)
4. 700 cm (too high)
5. 800 cm (too high)

Let's verify: If h = 600 cm, then total distance = \(3.625 \times 600 = 2175\) cm
This is very close to our target of 2,146 cm (difference of only 29 cm), which is reasonable given we're looking for an approximate answer.

Final Answer

The answer is C. 600 cm

Our calculation shows the original height was approximately 592 cm, and among the given choices, 600 cm is the closest match. The small difference between our exact calculation and the answer choice is expected since the problem asks for an approximate height.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what constitutes the "third time" hitting the ground
Students often get confused about counting the impacts. They might think the ball needs to complete three full bounce cycles instead of understanding that we stop counting distance when the ball hits the ground for the third time (after two bounces). This leads to including extra distance segments that shouldn't be counted.

2. Incorrectly mapping the ball's journey
Students frequently miss that each bounce involves TWO distance segments - going up to the bounce height AND coming back down from that same height. They might only count the upward bounces or only the downward falls, missing that both directions contribute to the total vertical distance traveled.

3. Confusion about what "rebounds 75 percent as high" means
Some students misinterpret this as the ball losing \(75\%\) of its height (keeping only \(25\%\)) rather than rebounding TO \(75\%\) of the previous height. This fundamental misunderstanding leads to using 0.25 instead of 0.75 as the bounce coefficient throughout their calculations.

Errors while executing the approach

1. Arithmetic errors in calculating successive bounce heights
When computing \(0.75 \times 0.75 = 0.5625\) for the second bounce height, students often make multiplication errors or decimal placement mistakes. These small errors compound when calculating the total distance, leading to significantly different final answers.

2. Incorrectly combining like terms when summing distances
When adding \(\mathrm{h} + 0.75\mathrm{h} + 0.75\mathrm{h} + 0.5625\mathrm{h} + 0.5625\mathrm{h}\), students might incorrectly group terms or make addition errors. For example, they might add \(0.75\mathrm{h}\) only once instead of twice, or miscalculate \(0.5625\mathrm{h} + 0.5625\mathrm{h} = 1.125\mathrm{h}\).

3. Division errors when solving for h
When calculating \(\mathrm{h} = 2146 \div 3.625\), students often struggle with dividing by a decimal. They might incorrectly convert this to a fraction or make calculation errors, arriving at values like 500-something instead of the correct 592.

Errors while selecting the answer

1. Choosing the mathematically exact answer instead of the closest approximation
Students calculate \(\mathrm{h} \approx 592\) cm correctly but then look for an answer choice of exactly 592 or get frustrated when they don't see it. They fail to recognize that since the problem asks for an "approximate height," they should select the closest available option (600 cm).

2. Second-guessing their calculation due to the gap between computed and answer choice values
When students see their calculated value of 592 cm doesn't exactly match 600 cm, they might assume they made an error and randomly pick a different answer choice instead of confidently selecting the closest one. The 8 cm difference seems significant to them, causing doubt about their methodology.