Problem 105 - College Algebra Essentials

Question

A basketball player's hang time is the time spent in the air when shooting a basket. The formula $$t=\frac{\sqrt{d}}{2}$$ models hang time, \(t,\) in seconds, in terms of the vertical distance of a player's jump, \(d,\) in feet. (image cannot copy) When Michael Wilson of the Harlem Globetrotters slamdunked a basketball, his hang time for the shot was approximately 1.16 seconds. What was the vertical distance of his jump, rounded to the nearest tenth of a foot?

Step-by-Step Solution

Answer

Michael Wilson jumped approximately \(2.7\) feet during his slam dunk shot.

Step 1: Identify the given values

In the problem, it is given that the player's hang time for the shot was approximately 1.16 seconds. So, \(t = 1.16\) seconds.

Step 2: Rewrite the formula

Given the formula for hang time \(t = \frac{\sqrt{d}}{2}\), we want to solve for \(d\), the vertical distance of the jump. To do this, isolate \(d\) by multiplying both sides by 2 then squaring both sides, to get \(d = (2t)^2\). This will allow us to solve for \(d\) easily.

Step 3: Substitute and Solve

By substituting the given value of \(t\) into the new formula, we have \(d = (2*1.16)^2\). Calculate this value to find the distance \(d\).

Step 4: Round to the nearest tenth

Round the calculated distance to the nearest tenth of a foot as asked by the problem.

Key Concepts

Mathematical Modeling

Mathematical modeling is the process where real-life situations are represented using mathematical expressions or formulas. In this exercise, the hang time of a basketball player is modeled using a specific formula:

• The formula given is: \[ t=\frac{\sqrt{d}}{2} \]
• "t" represents the hang time in seconds.
• "d" symbolizes the vertical distance in feet.

This mathematical representation helps us understand and predict the hang time based on the height someone jumps. By converting a physical situation to a mathematical context, we can solve problems systematically. This formula associates the player's hang time with the square root of the vertical distance, divided by two. Understanding and creating these types of models are crucial in applying math to real-word scenarios.

Algebraic Manipulation

Algebraic manipulation is a key math skill using basic algebraic rules to rearrange equations and solve for variables. In our exercise, we start with the hang time formula and solve for the vertical distance "d". The steps include:

• Start with the formula: \( t = \frac{\sqrt{d}}{2} \).
• To isolate \( \sqrt{d} \), multiply both sides by 2, resulting in \( 2t = \sqrt{d} \).
• Square both sides to remove the square root: \( (2t)^2 = d \).

These manipulations allow us to find "d", the desired variable, by reorganizing the given formula. It's crucial to perform each algebraic step carefully to avoid mistakes. This process teaches us how to transform an initial equation to achieve the desired expression or solution.

Problem Solving

Problem solving involves a series of steps to find a solution to a given question or scenario. Let's break down how we tackled this particular problem:

• First, identify the known values in the problem. Here, the hang time \( t = 1.16 \) seconds is given.
• Next, alter the formula mathematically to solve for the unknown, \( d \), using algebraic manipulation.
• Substitute the known value into the newly acquired formula: \( d = (2 \times 1.16)^2 \).
• Calculate the expression to obtain your solution: \( d = 5.3824 \).
• Finally, round the result to match the problem's requirement: approximately 5.4 feet.

These steps demonstrate the logical order needed to approach complex problems methodically, ensuring the correct application of mathematics to find an accurate solution.

Vertical Distance Calculation

Vertical distance calculation is crucial in many areas, particularly in physics and sports, as demonstrated in this problem. The need to find the jump height, given a known hang time, makes understanding this process important:

• This exercise uses a known relationship between the time in the air and the distance jumped.
• Substitution in the derived formula provides specific figures: substituting \( t \) = 1.16 gives us \( d = (2 \times 1.16)^2 \).
• The result, \( d = 5.3824 \), refers to the height traveled in feet.
• Rounding is used to align with the context - here, we round 5.3824 to 5.4 feet.

Vertical distance calculations, like these, require clear understanding of the physical conditions and precise use of mathematical formulas, ensuring we arrive at accurate and meaningful results.