Problem 67 - Intermediate Algebra

Question

The height of a projectile fired vertically into the air (neglecting air resistance) at an initial velocity of 64 feet per second is a function of the time \((t)\) and is given by the equation $$ h(t)=64 t-16 t^{2} $$ Compute \(h(1), h(2), h(3)\), and \(h(4)\).

Step-by-Step Solution

Answer

\( h(1) = 48 \), \( h(2) = 64 \), \( h(3) = 48 \), \( h(4) = 0 \) feet.

Step 1: Identify the function

The height of the projectile at time \( t \) is represented by the function \( h(t) = 64t - 16t^2 \). This function will be used to find the height at specific times.

Step 2: Calculate \( h(1) \)

Substitute \( t = 1 \) into the equation \( h(t) = 64t - 16t^2 \). \[ h(1) = 64(1) - 16(1)^2 = 64 - 16 = 48 \] Thus, \( h(1) = 48 \) feet.

Step 3: Calculate \( h(2) \)

Substitute \( t = 2 \) into the equation \( h(t) = 64t - 16t^2 \). \[ h(2) = 64(2) - 16(2)^2 = 128 - 64 = 64 \] Thus, \( h(2) = 64 \) feet.

Step 4: Calculate \( h(3) \)

Substitute \( t = 3 \) into the equation \( h(t) = 64t - 16t^2 \). \[ h(3) = 64(3) - 16(3)^2 = 192 - 144 = 48 \] Thus, \( h(3) = 48 \) feet.

Step 5: Calculate \( h(4) \)

Substitute \( t = 4 \) into the equation \( h(t) = 64t - 16t^2 \). \[ h(4) = 64(4) - 16(4)^2 = 256 - 256 = 0 \] Thus, \( h(4) = 0 \) feet, indicating the projectile has returned to the launch height.

Key Concepts

Quadratic Functions

Quadratic functions are a type of polynomial function characterized by the highest degree being 2. This makes their equations look like a parabola when graphed. For our projectile motion exercise, the quadratic function is written as:\[ h(t) = 64t - 16t^2 \]This parabola opens downwards because the coefficient of the squared term, \(t^2\), is negative. Quadratic functions are crucial in projectile motion because they model how the height changes over time under constant acceleration due to gravity.
In simple terms, this equation shows how the height of a projectile changes based on time. The projectile goes up and reaches a maximum point; afterward, it comes back down. The graph will show a peak at the maximum height before descending.

• The graph of a quadratic function is a "U" shaped curve called a parabola.
• The point where the parabola changes direction is called the vertex.
• In projectile motions, the vertex represents the maximum height of the object.

Understanding quadratic functions helps in anticipating how high or far a projectile will go within a given timeframe.

Initial Velocity

Initial velocity is the speed at which an object starts its motion. In our case, the projectile is launched with an initial velocity of 64 feet per second. This value is critical because it determines how high and how quickly the projectile will move upward initially.
A higher initial velocity means the object will travel higher and stay in the air longer before gravity pulls it back down. The initial velocity directly affects the equation of motion, especially the linear term:\[ 64t \]In physics, the initial velocity is important for predicting the behavior of any projectile. The initial velocity provides the starting speed and direction, representing the object's energy moving it upwards.

• Initial velocity determines the height and distance of projectile travel.
• It is the coefficient of the linear term in the height-time function.
• The greater the initial velocity, the higher the projectile can go.

Thus, mastering the concept of initial velocity allows one to better evaluate the outcomes of a projectile's journey.

Time and Height Relationship

The relationship between time and height in the context of the given quadratic function is a core feature of projectile motion. As time progresses, the height changes according to the function:\[ h(t) = 64t - 16t^2 \]This equation reveals that height (h) is dependent on the time (t) since the projectile began its flight.
Initially, as time increases, height increases until maximum height is achieved. Afterwards, height decreases as the object returns to the ground. These changes are calculated by substituting different values of \(t\) into the equation.

• For \(t = 1\), \(h(1) = 48\) feet, showing an increase in height.
• For \(t = 2\), \(h(2) = 64\) feet, showing maximum height.
• For \(t = 3\), \(h(3) = 48\) feet, height decreases as it descends.
• For \(t = 4\), \(h(4) = 0\) feet, the projectile returns to the initial height.

Understanding this relationship helps predict when and where a projectile will be at various times, crucial for applications like ballistics, sports, and engineering projects.