Problem 69 - Precalculus: Mathematical for Calculus

Question

A ball is dropped from a height of \(9 \mathrm{ft}\). The elasticity of the ball is such that it always bounces up one-third the distance it has fallen. (a) Find the total distance the ball has traveled at the instant it hits the ground the fifth time. (b) Find a formula for the total distance the ball has traveled at the instant it hits the ground the \(n\) th time.

Step-by-Step Solution

Answer

(a) The ball travels approximately 18.85 ft. (b) \( D_n = 9 + 6(1 - (\frac{1}{3})^{n-1}) \).

Step 1: Initial setup and first drop

Initially, the ball is dropped from a height of 9 ft. This is the first drop, so the ball travels 9 ft downwards to the ground on its first fall.

Step 2: First bounce

Upon hitting the ground, the ball bounces back to one-third of the height it fell from. Therefore, it bounces back to a height of \( \frac{9}{3} = 3 \) ft. Then, it falls the same 3 ft back to the ground.

Step 3: Second bounce

After falling back to the ground for the second time, it bounces back to \( \frac{1}{3} \) of 3 ft, which is 1 ft. Then, it falls 1 ft back to the ground.

Step 4: Third bounce

The ball bounces again to \( \frac{1}{3} \) of 1 ft, which is \( \frac{1}{3} \) ft, and then falls \( \frac{1}{3} \) ft.

Step 5: Fourth and Fifth bounce

Repeat the same process: the ball bounces \( \frac{1}{3} \) of \( \frac{1}{3} \) ft = \( \frac{1}{9} \) ft and falls \( \frac{1}{9} \) ft. Then it bounces \( \frac{1}{3} \) of \( \frac{1}{9} \) ft = \( \frac{1}{27} \) ft and falls \( \frac{1}{27} \) ft on its fifth landing.

Step 6: Calculate total distance for first five ground touches

Add the distances: the initial 9 ft fall, the sum of all the bounce and fall heights: - 1st bounce: \( 3 + 3 = 6 \) ft - 2nd bounce: \( 1 + 1 = 2 \) ft - 3rd bounce: \( \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \) ft - 4th bounce: \( \frac{1}{9} + \frac{1}{9} = \frac{2}{9} \) ft - 5th bounce: \( \frac{1}{27} \) ft (only upward) Total = \( 9 + 6 + 2 + \frac{2}{3} + \frac{2}{9} + \frac{1}{27} \approx 18.85 \) ft.

Step 7: Develop formula for nth ground touch

Notice the pattern: the total distance after \( n \) bounces consists of the first fall and a geometric series for subsequent bounces. The formula for the total distance after the nth ground touch is:\[ D_n = 9 + 2 \times \sum_{k=1}^{n-1} \left(9 \left(\frac{1}{3}\right)^k\right) \] Simplifying, using the sum formula for geometric series, we get: \[ D_n = 9 + 2 \times \left( \frac{3}{2}(1 - (\frac{1}{3})^{n-1}) \right) \]

Key Concepts

Geometric Series

A geometric series is a sum of numbers each of which is a constant multiple of the previous number. This constant multiple is called the "common ratio."
In the context of the bouncing ball problem, observing the subsequent bounces can help us identify a geometric series form.
The ball bounces back up to a height, consistently multiplying the prior distance by the common ratio of \(\frac{1}{3}\).

• First bounce: \(3\) ft
• Second bounce: \(1\) ft
• Third bounce: \(\frac{1}{3}\) ft
• Fourth bounce: \(\frac{1}{9}\) ft

We can define a geometric series with each bounce as \(a_r^k\), where \(a\) is the initial drop and \(r\) the common ratio. This makes a powerful tool for modeling repetitive, multiplying events like this.
The total distance calculation in this problem includes the sum of a geometric series, determined by: \[S_n = a_1 \frac{1 - r^n}{1 - r}\]This formula aids in calculating the cumulative height of all bounces. Recognizing geometric series in problems is key to solving such problems effectively.

Recursive Sequences

Recursive sequences provide a means to describe a process in terms of itself. In other words, they establish each term based on the previous one, often through some repetitive transformation.
In our ball scenario, every bounce that follows can be expressed as a recursive sequence due to its reliance on the bounce before it.

• Initial height (first term): \(9\) ft
• Recursive relation for bounces: \(h_{n+1} = \frac{1}{3} h_n\)

Where \(h_{n+1}\) is the height of the \(n+1\)-th bounce and \(h_n\) is the height of the \(n\)-th bounce.
This iterative pattern shows the ball's height continuously shrinking by a factor of \(\frac{1}{3}\) every subsequent bounce.
Recursive sequences emphasize the essence of capturing processes that naturally develop over time, step by step.

Mathematical Problem Solving

Mathematical problem solving is at the heart of understanding and tackling real-world scenarios by utilizing various mathematical concepts and techniques.
In the bouncing ball exercise, multiple mathematical strategies come into play:

• Identifying patterns: Recognizing the repetitive behavior of the ball through geometric sequences and recursive sequences.
• Generalizing: Deriving a formula that can apply broadly to any number of bounces \(n\).
• Utilizing formulas: Applying sum formulas of geometric series to simplify the explanation of the ball’s motion.

Effective problem solving involves:

• Breaking down complex problems into simpler steps, as demonstrated in sequential bounce calculations.
• Applying known mathematical formulas and concepts to new problems.
• Thinking creatively to connect various concepts like sequences and series to reach a solution.

Through practice and comprehensive understanding, the problem-solving process becomes a powerful technique for tackling unfamiliar mathematical challenges, equipping students with skills to explore complex systems and discover their underlying patterns.