Problem 57 - Beginning and Intermediate Algebra

Question

A model rocket is projected vertically upward from the ground. Its distance \(s\) in feet above the ground after t seconds is given by the quadratic function $$ s(t)=-16 t^{2}+256 t $$ Work Exercise in order, to see how quadratic equations and inequalities are related. At what times will the rocket be at ground level? (Hint: Let \(s(t)=0\) and solve the quadratic equation.)

Step-by-Step Solution

Answer

The rocket will be at ground level at \( t = 0 \) and \( t = 16 \) seconds.

Step 1 - Identify the equation

The height of the rocket as a function of time is given by the quadratic equation: \[ s(t) = -16t^2 + 256t \]

Step 2 - Set the height to zero

To find when the rocket will be at ground level, set the height function to zero: \[ 0 = -16t^2 + 256t \]

Step 3 - Factor the equation

Factor out common terms from the equation: \[ 0 = -16t(t - 16) \]

Step 4 - Solve for t

Set each factor to zero and solve for the values of \( t \): \[ -16t = 0 \quad \text{or} \quad t - 16 = 0 \] Thus, \( t = 0 \) and \( t = 16 \).

Step 5 - Interpret the solution

The rocket will be at ground level at two times: initially at \( t = 0 \) seconds and again at \( t = 16 \) seconds.

Key Concepts

Vertical Motion

When studying the behavior of objects projected upwards, we often encounter vertical motion. This type of motion is described by quadratic equations, which help us understand how the object moves. In this exercise, the model rocket's height above the ground over time follows the equation: \[ s(t) = -16t^2 + 256t \]This equation tells us how high the rocket will be after any given time 't'. The term \[-16t^2\] represents the effect of gravity pulling the rocket downwards, while \[256t\] accounts for the initial velocity pushing it up. These two terms work together to describe the rocket's vertical path.

Factoring

Factoring is a method used to simplify quadratic equations, making them easier to solve. In our rocket example, we want to know when the rocket is at ground level. We start with the equation: \[0 = -16t^2 + 256t\]We notice that each term has a common factor: - The number '16' can be factored out since each coefficient is divisible by 16. - The variable 't' is present in both terms. By factoring out the common terms, we get: \[0 = -16t(t - 16)\]This simplifies our equation and makes solving for 't' much easier. Essentially, factoring breaks down the problem into more straightforward parts.

Solving Quadratic Equations

After factoring the equation, we can solve for the values of 't'. From our factored equation: \[0 = -16t(t - 16)\]We set each factor equal to zero: \[-16t = 0\] or \[t - 16 = 0\]Solving these, we find: \[t = 0\] and \[t = 16\]This process tells us the times at which the rocket will be on the ground. So the rocket starts on the ground at \[t = 0\] seconds and returns to the ground at \[t = 16\] seconds. Solving quadratic equations in this way helps us pinpoint critical moments in the object's motion.

Projectile Motion

In this exercise, we are dealing with projectile motion, a type of motion experienced by objects that are thrown or projected into the air. The path these objects follow is affected by gravity, which pulls them back towards the earth. The equation \[ s(t) = -16t^2 + 256t \] models the vertical path of our rocket. Key points to understand about projectile motion: - The object will rise to a peak height and then fall back down.- Quadratic equations are used to determine critical points in its trajectory, such as when it hits the ground.- The initial velocity plays a significant role in how high and how far the object will travel. By using quadratic equations, we can accurately predict the path of any projectile, which is extremely useful in fields like physics and engineering.