Problem 54 - A Graphical Approach to College Algebra

Question

Suppose that when a ball is dropped, the height of its first rebound is about \(80 \%\) of the initial height that it was dropped from, the second rebound is about \(80 \%\) as high as the first rebound, and so on. If this ball is dropped from 12 feet in the air, model the height in feet of each rebound with an exponential function \(H(x),\) where \(x=0\) represents the initial height, \(x=1\) represents the height on the first rebound, and so on. Find the height of the third rebound. Determine which rebound had a height of about 2.5 feet.

Step-by-Step Solution

Answer

Third rebound: 6.144 feet; 2.5 feet height: 8th rebound.

Step 1: Understanding the Problem

We are given a scenario where a ball rebounds to 80% of its previous height each time it bounces. Initially, the ball is dropped from 12 feet. We need to model this scenario with an exponential function and find specific rebound heights.

Step 2: Defining the Exponential Function

The rebound height can be described by an exponential function because it reduces by a constant percentage each time. The general form of the function is: \[ H(x) = H_0 \times r^x \]where \(H_0\) is the initial height (12 feet) and \(r\) is the rebound ratio (0.8). Thus, the function becomes: \[ H(x) = 12 \times 0.8^x \].

Step 3: Calculating the Third Rebound Height

Substitute \(x = 3\) into the function to find the height after the third rebound: \[ H(3) = 12 \times 0.8^3 \].Calculate \(0.8^3\), which is 0.512. Then calculate the height:\[ H(3) = 12 \times 0.512 = 6.144 \]feet.

Step 4: Determining which Rebound has 2.5 Feet Height

We need to solve \(H(x) = 2.5\) to find which rebound has this height:\[ 12 \times 0.8^x = 2.5 \].Dividing both sides by 12 gives:\[ 0.8^x = \frac{2.5}{12} \approx 0.2083 \].Take the natural logarithm of both sides:\[ \ln(0.8^x) = \ln(0.2083) \].Apply the logarithm power rule \((\ln(a^b) = b\ln(a))\):\[ x \ln(0.8) = \ln(0.2083) \].Now, solve for \(x\):\[ x = \frac{\ln(0.2083)}{\ln(0.8)} \approx 7.48 \].Since \(x\) must be a whole number, round 7.48 upwards to 8. Therefore, the 8th rebound is about 2.5 feet.

Key Concepts

Rebound Ratio

In the context of a bouncing ball, the rebound ratio is a simple yet crucial concept. It tells us how high the ball rebounds relative to the height from which it was dropped. In mathematical terms, this ratio remains constant for each bounce. In our example, the rebound ratio is 80% or 0.8. This means that after each bounce, the ball reaches 80% of the height it was at before. Understanding the rebound ratio helps in predicting the ball's behavior over multiple bounces. It's a constant that you multiply the previous height by to find the height of the next bounce. If a ball rebounds to 0.8 of its height each time, then: - The first bounce height is 0.8 times the original drop height. - The second bounce height is 0.8 times the first bounce height. By applying this concept repeatedly, you can model the ball's motion and predict each rebound height.

Exponential Decay

Exponential decay is a common mathematical process in which a quantity decreases at a rate proportional to its current value. In our bouncing ball scenario, each bounce represents a step in an exponential decay process.The formula used to describe this type of decay is:\[ H(x) = H_0 \times r^x \]This represents how the ball's height changes over time, where:

• \(H(x)\) is the height at the \(x\)-th bounce.
• \(H_0\) is the initial drop height.
• \(r\) is the fixed rebound ratio (0.8 in our case).
• \(x\) is the number of bounces.

Every time the ball bounces, its height decreases according to this exponential function. For example, the height after three bounces turns out to be \(6.144\) feet when calculated using \[ H(3) = 12 \times 0.8^3 \]. This highlights how exponential decay efficiently models the height changes over multiple bounces.

Mathematical Modeling

Mathematical modeling is the process of representing real-world situations through mathematical equations. It allows us to make predictions and analyze behaviors without physical trials. Here, we use an exponential function to model the bouncing ball scenario.To start, consider the initial condition: the ball is dropped from a certain height. Each bounce thereafter is a representation of the decay due to the rebound ratio. By setting up our model using \[ H(x) = 12 \times 0.8^x \], we can assess each bounce's height. For practical applications like determining which specific bounce results in a height of 2.5 feet, the model helps us solve for \(x\). We do this by setting \[ H(x) = 2.5 \] and solving the equation: \[ 12 \times 0.8^x = 2.5 \]. Solving tells us that around the 8th rebound, the ball reaches approximately 2.5 feet.Mathematical modeling like this provides an insight into dynamic systems, predict outcomes, and understand underlying mechanisms.