Problem 57 - College Algebra with Corequisite Support

Question

For the following exercises, determine the function described and then use it to answer the question. An object dropped from a height of 600 feet has a height, \(h(t),\) in feet after \(t\) seconds have elapsed, such that \(h(t)=600-16 t^{2}\). Express \(t\) as a function of height \(h,\) and find the time to reach a height of 400 feet.

Step-by-Step Solution

Answer

Time to reach 400 feet is approximately 3.54 seconds.

Step 1: Express t as a function of h

Start with the given function for height: \[ h(t) = 600 - 16t^2 \]. We need to express \( t \) as a function of \( h \). First, solve the equation for \( t^2 \):\[ 16t^2 = 600 - h \]. Now divide by 16:\[ t^2 = \frac{600 - h}{16} \]. Take the square root of both sides to solve for \( t \):\[ t = \sqrt{\frac{600 - h}{16}} \].

Step 2: Substitute h = 400 to find t

Now that we have \( t(h) = \sqrt{\frac{600 - h}{16}} \), we will substitute \( h = 400 \) to find the time \( t \) at which the height is 400 feet:\[ t = \sqrt{\frac{600 - 400}{16}} \]. This simplifies to:\[ t = \sqrt{\frac{200}{16}} \].

Step 3: Simplify the expression

Simplify the expression from Step 2:\[ t = \sqrt{12.5} \].Solve for \( t \):\[ t = \sqrt{12.5} \approx 3.54 \].

Step 4: Verify the time solution

Verify by substituting \( t \approx 3.54 \) back into the height function:\[ h(3.54) = 600 - 16 \times (3.54)^2 \].Calculate \( 16 \times 3.54^2 \) and verify that it approximately equals 200:\( 16 \times 3.54^2 \approx 200 \), thus confirming the solution.

Key Concepts

Quadratic Functions

A quadratic function is a type of polynomial function where the variable is raised to the power of 2. It can generally be expressed in the form \( f(x) = ax^2 + bx + c \). This type of function creates a parabola when graphed on a coordinate plane.
Quadratic functions have several important features:

• **Vertex:** The highest or lowest point of the parabola.
• **Axis of Symmetry:** A vertical line that divides the parabola into two mirror images.
• **Roots or Zeros:** The values of \( x \) for which \( f(x) = 0 \).
• **Direction:** Parabolas open upwards if \( a > 0 \) and downwards if \( a < 0 \).

In solving quadratic problems like our original exercise, recognizing the function's structure allows us to manipulate and solve it for different variables, such as solving for \( t \) in terms of \( h \) or vice versa.

Solving Equations

Solving equations involves finding the value of the variable that makes the equation true. In our context, we are transforming a quadratic function to express one variable in terms of the other.
Here's how we approached solving the equation:

• Starting with the given equation \( h(t) = 600 - 16t^2 \), we aim to express \( t \) as a function of \( h \).
• We isolated \( t^2 \) by rearranging the equation to \( 16t^2 = 600 - h \).
• Next, we reduced the equation further by dividing both sides by 16, simplifying it to \( t^2 = \frac{600 - h}{16} \).
• Finally, we took the square root of both sides, giving us \( t = \sqrt{\frac{600 - h}{16}} \).

This process demonstrates solving a quadratic equation step by step by isolating the variable and simplifying the expression.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging equations to isolate variables or simplify expressions. It's a fundamental aspect of solving equations and involves operations such as addition, subtraction, multiplication, division, and taking roots.
In the exercise, we performed algebraic manipulation to express \( t \) as a function of \( h \):

• We rearranged the original equation \( h(t) = 600 - 16t^2 \) to isolate terms involving \( t \).
• By dividing both sides by 16, an important algebraic step, we simplified the expression.
• Finally, applying the square root to simplify further and solve for \( t \), acknowledging that \( t \) must be real and positive as it represents time.

Such manipulations allow us to effectively handle functions and equations, providing necessary transformations to obtain desired results or solutions.