Problem 51 - College Algebra

Question

Use the position formula $$ s=-16 t^{2}+v_{0} t+s_{0} $$ \(\left(v_{0}=\text { initial velocity, } s_{0}=\text { initial position, } t=\text { time }\right)\) to answer Exercises \(49-52 .\) If necessary, round answers to the nearest hundredth of a second. A ball is thrown vertically upward with a velocity of 64 feet per second from the top edge of a building 80 feet high. For how long is the ball higher than 96 feet?

Step-by-Step Solution

Answer

The ball is higher than 96 feet for approximately 3.46 seconds.

Step 1: Identify Knowns

Identify the known values from the problem statement: \(v_{0}\) (initial velocity) = 64 feet/second, \(s_{0}\) (initial position) = 80 feet, and the height the ball needs to be higher than (96 feet).

Step 2: Set up the Equation

Set up the position formula with the target height on the left and the position equation on the right. This will look like: 96 = -16\(t^{2}\) + 64t + 80.

Step 3: Solve the Quadratic Equation

To find the values of 't' when 's' is 96, rearrange the equation into a quadratic form: -16\(t^{2}\) + 64t + 80 - 96 = 0. This simplifies to -16\(t^{2}\) + 64t -16 = 0. To solve for 't', divide the whole equation by -16 to get: \(t^{2}\) - 4t +1 = 0.

Step 4: Determine the Solution for the Quadratic

Using the quadratic formula \(t = [-b \pm sqrt{(b^{2}-4ac)}]/2a\), the solution set for 't' can be determined, where a=1, b=-4 and c = 1. Plugging these in, we find \(t = [4 \pm sqrt{(-4)^{2}-4(1)(1)}]/2(1)\), which simplifies to \(t = [2 \pm sqrt{3}]\) or approximately t = 0.27 seconds and t = 3.73 seconds.

Step 5: Interpret the Results

The two values for 't' represent the two occasions when the ball is at a height of 96 feet during its trajectory; once while on its way up, and again when it's coming down. By subtracting the smaller time from the larger one, we can find the duration that the ball is above 96 feet. So, 3.73s - 0.27s = approximately 3.46 seconds.

Key Concepts

Quadratic Equations

At the heart of many algebraic problems lies the quadratic equation, a type of polynomial equation of the second degree. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are coefficients, and \(a \eq 0\). In the context of projectile motion, for example, a quadratic equation can describe how the height of an object changes over time.

Understanding quadratic equations is crucial because they pop up in various areas of math and science. They are used to find optimal solutions, calculate areas, and even in business for profit maximization. The solutions to these equations, often found by factoring, completing the square, or using the quadratic formula, provide essential information about the graph of the quadratic function, namely the location of its vertex and its x-intercepts, or roots.

Initial Velocity

The term 'initial velocity' refers to the speed of an object at the start of its motion. In projectile motion problems, the initial velocity is a critical component, as it sets the stage for how the object will behave as it moves. If you know the initial velocity and the forces acting on the object, like gravity or air resistance, you can predict its trajectory.

Initial velocity is not just a number; it's a vector quantity, which means it has both magnitude (speed) and direction. For example, in our exercise, the initial velocity is upward at 64 feet per second from the building's edge. This initial information helps paint the full picture of the object's journey through space and is essential for solving related algebraic problems using the position formula.

Projectile Motion in Algebra

Projectile motion refers to the motion of an object that is thrown or launched into the air and is subject to gravity's acceleration. Algebraically, it's often modeled by quadratic equations reflecting the object's vertical motion. The position formula \( s = -16t^2 + v_{0}t + s_{0} \) demonstrates this, where \( s \) stands for the object's position at any given time \( t \) where \( t \geq 0 \), \( v_{0} \) denotes the initial velocity, and \( s_{0} \) the initial position.

In the language of algebra, the equation accounts for the predictable way gravity influences vertical movement: as an acceleration that consistently pulls the object downward, thereby affecting its height over time. This downward force is represented by the constant -16 feet per second² in the position formula, typical for objects in freefall near Earth's surface.

Solving Quadratic Equations

Solving quadratic equations can be approached in several ways, with the quadratic formula being one of the most powerful and universal methods. Given \( ax^2 + bx + c = 0 \), the quadratic formula \( x = [-b \pm \sqrt{b^2-4ac}]/2a \) allows us to find the values of \( x \) for any quadratic equation. When we apply this formula, we're looking for the roots of the equation, which are the \( x \) values that make the equation true.

Using the quadratic formula often involves simplifying the equation first, as we did by factoring out a common term in our example problem. Once the equation is in its simplest form, the formula can quickly supply answers that might otherwise require more complex factoring. The \(+/-\) sign indicates that there will generally be two solutions, which makes sense when considering the symmetrical shape of a parabolic graph, touching the \( x \) axis at most two points—or not at all if the solutions are complex numbers.