Problem 20 - College Algebra Essentials

Question

A football was kicked vertically upward from a height of 4 feet with an initial speed of 60 feet per second. The formula $$h=4+60 t-16 t^{2}$$ describes the ball's height above the ground, \(h,\) in feet, \(t\) seconds after it was kicked. Use this formula to solve Exercises \(19-20 .\) What was the ball’s height 3 seconds after it was kicked?

Step-by-Step Solution

Answer

The ball's height 3 seconds after it was kicked is 40 feet.

Step 1 - Understanding the Problem

The question is asking for the height of the football 3 seconds after it was kicked. This can be found by substituting t=3 into the height formula given, \( h = 4 +60t -16t^2 \). The variable 't' in the formula represents the time in seconds after the kick, and the 'h' variable represents the height in feet. Hence 't' = 3 needs to be substituted in this formula to calculate the height after 3 seconds.

Step 2 - Substituting the value of t into the formula

The value 't' = 3 is substituted into the formula: \( h = 4 +60*3 -16*3^2 \).

Step 3 - Simplifying the equation

Firstly, solve the multiplication and then the addition and subtraction from left to right: \( h = 4+180-144\)

Step 4 - Calculating the final height

Solving the above, we have a final equation: \( h = 40 \)

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental component of algebra and appear frequently in various mathematical applications, including vertical motion problems. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. These equations create a parabolic shape when graphed, either opening upwards or downwards, depending on the sign of \( a \). In the context of our problem, the football's height is described by a quadratic equation: \( h = 4 + 60t - 16t^2 \), where:

• \( 4 \) is the initial height in feet,
• \( 60t \) represents the initial velocity component, and
• \( -16t^2 \) accounts for the effect of gravity, a common coefficient in equations representing objects in free fall near the Earth's surface.

To solve such equations, we typically substitute a specific value for the variable \( t \) (time, in this activity) and simplify the expression to find \( h \), the height. Understanding and working with quadratic equations is key to solving many physics and algebra problems effectively.

Physics Applications in Algebra

Physics often uses algebraic expressions to represent physical phenomena, which allows us to model and solve real-world problems. One common application is in modeling vertical motion, as seen with the football example. In this scenario, algebra helps link the physical principles of initial velocity and the acceleration due to gravity to mathematical expressions.The equation \( h = 4 + 60t - 16t^2 \) is a perfect illustration:

• The constant term \( 4 \) represents the ball's initial height above the ground.
• The term \( 60t \) depicts the contribution of initial velocity. Here, \( 60 \) feet per second is the speed at which the ball was initially kicked vertically.
• \( -16t^2 \) represents the effect of Earth's gravity, which causes the upward motion to slow down and eventually reverse.

By replacing \( t \) with specific time values, we determine the ball's height at those moments. This helps us understand and predict the ball's trajectory, demonstrating the powerful interplay between physics and algebra.

Evaluating Algebraic Expressions

Evaluating algebraic expressions means calculating the value of expressions for given values of variables. This involves substituting the variables with actual numbers and performing arithmetic operations following the order of operations (PEMDAS/BODMAS).In our football problem, we aim to find the ball's height \( h \) 3 seconds after being kicked. Let's break it down:

• First, substitute \( t = 3 \) into the formula: \( h = 4 + 60(3) - 16(3)^2 \).
• Next, calculate step-by-step: \( 60(3) = 180 \) and \( 16(3)^2 = 144 \), making the expression \( h = 4 + 180 - 144 \).
• Finally, combine the simplified terms: \( h = 40 \).

This illustrates the process of evaluating expressions to solve real-world problems. Mastering this skill is crucial in both algebra and practical applications, as it allows you to find answers from given mathematical models.