Problem 77 - Precalculus Mathematics 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 onethird 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

The total distance after the fifth hit is \(\frac{485}{27}\) ft. For the nth hit, the distance is \(18 - 9\left(\frac{1}{3}\right)^n \) ft.

Step 1: Understanding Initial Conditions

When the ball is dropped, it falls a distance of 9 feet initially. This is our starting point, as the ball must first drop this distance before it begins to bounce.

Step 2: Calculate the First Bounce

After hitting the ground, the ball bounces up to a height that is one third of the 9 feet it initially fell, which is \( \frac{9}{3} = 3 \) feet. It then falls back another 3 feet to the ground.

Step 3: Calculate Second Bounce

The ball now falls from a height of 3 feet and bounces to one third of this height, which is \( \frac{3}{3} = 1 \) foot. It falls another 1 foot back to the ground.

Step 4: Calculate Third Bounce

From the 1 foot height, it bounces to \( \frac{1}{3} = \frac{1}{3} \) foot and falls back this \( \frac{1}{3} \) foot.

Step 5: Calculate Fourth and Fifth Bounces

Continuing this pattern, on the fourth bounce, the ball rises to \( \frac{1}{9} \) foot and falls back the same distance. On the fifth bounce, it rises to \( \frac{1}{27} \) foot and falls back.

Step 6: Sum Up All Distances for Five Hits

The total distance traveled is the sum of all the falling and bouncing distances. This includes the initial 9 feet drop, and then, for each bounce, twice the height bounced: \(9 + (3 + 3) + (1 + 1) + \left(\frac{1}{3} + \frac{1}{3}\right) + \left(\frac{1}{9} + \frac{1}{9}\right) + \left(\frac{1}{27} + \frac{1}{27}\right).\)

Step 7: Calculate the Total Distance for Five Hits

Adding these together, we get:\(9 + 6 + 2 + \frac{2}{3} + \frac{2}{9} + \frac{2}{27}.\)Converting to a common fraction base, the sum is:\( 9 + 6 + 2 + \frac{6}{9} + \frac{2}{9} + \frac{2}{27} = 17 + \frac{8}{9} + \frac{2}{27}.\)Simplify to: \(17 + \frac{24}{27} + \frac{2}{27} = 17 + \frac{26}{27} = \frac{459}{27} + \frac{26}{27} = \frac{485}{27}.\)

Step 8: Establish a General Formula for nth Bounce

Since each subsequent bounce is \(\frac{1}{3}\) of the previous bounce height, we can create a general formula by recognizing this geometric progression. The sum of both falling and bouncing legs of each journey adds up to:\(D_n = 9 + 2\left(3 \left(\frac{1 - \left(\frac{1}{3}\right)^n}{1 - \frac{1}{3}}\right) \right) = 9 + 2 \left(4.5 \left(1 - \left(\frac{1}{3}\right)^n \right)\right).\)

Step 9: Present the Simplified Formula

Simplify the formula to:\[D_n = 9 + 9 \left(1 - \left(\frac{1}{3}\right)^n \right) = 18 - 9\left(\frac{1}{3}\right)^n .\]

Key Concepts

Bounce Distance

When a ball is dropped from a certain height, it exhibits a fascinating behavior called bounce distance. This is the distance it travels after every bounce. Initially, when a ball is dropped from 9 feet, it forgets nothing of its high jump. As soon as it touches the ground for the first time, it bounces back up to one-third of its previous height. So, it initially goes up to 3 feet. Then, it has to do a little more work and fall back the same 3 feet to hit the ground again.

Every bounce reduces the upward height to one-third of the previous height. For example, the second bounce, which starts from 3 feet, reaches only 1 foot high before descending back, and each successive bounce continues this pattern. It's as if the ball is a master in the art of shrinking expectations! Each round trip (up then back down) for bouncing is the distance traveled for that bounce. All these distances add up in a sequence that involves both the heights it rises to and falls back from. Understanding bounce distance helps unravel amazing numerical patterns in sequences.

Series Summation

Summing up all the distances the ball travels might at first seem tricky, but fear not, for series summation comes to the rescue! This geometric sequence starts with an initial height and keeps decreasing according to the elasticity rule. The initial drop of the ball was 9 feet. But every time after that, when it bounces, this bounce happens in pairs: going up and then coming down.

To find the total distance the ball has traveled by the fifth bounce, we perform a summation. The total distance includes the initial drop of 9 feet and twice the amount of each bounce height thereafter. For instance, after its first drop, the sequence continues as 3 feet up and down, then 1 foot up and down, and further continues with smaller sections like \( \frac{1}{3} \) and \( \frac{1}{9} \) feet each time eventually making it a streak of diminishing numbers.

With the geometric series formula, we add up these components:

• The initial distance (9 feet)
• Twice the sum of the infinite sequence (since each bounce up and down is counted twice)

The series allows us to encapsulate all these distances into a single formula, saving both time and pen strokes!

Mathematical Modeling

Mathematical modeling transforms this bouncing ball problem into a manageable formula. This formula represents the sum of the bounce series, each bounce’s height being one-third of the last. This transformation simplifies and predicts behavior, exactly the kind of capability one would hope for in mathematics!

By creating a general formula, you can find the bounce distance for any number of bounces the ball makes. This model considers that after each contact with the ground, the ball rises again, only to a smaller extent. This journey of up and down, multiplying each time by the elasticity factor \( \frac{1}{3} \), turns into a smart formula: \[ D_n = 18 - 9\left(\frac{1}{3}\right)^n \] where \( D_n \) stands for the total distance traveled by the \( n \)th bounce iteration.

This model is a compact summary that holds the secrets of infinite repeating patterns within finite terms. Such mathematical ingenuity not only impresses but also saves much computational fumbling around, conveying the power of algebra in describing actions as common yet captivating as a bouncing ball!