Problem 35 - Algebra 2

Question

PHYSICS For Exercises \(34-36,\) use the following information. An object is fired straight up from the top of a 200 -foot tower at a velocity of 80 feet per second. The height \(h(t)\) of the object \(t\) seconds after firing is given by \(h(t)=-16 t^{2}+80 t+200\) Find the maximum height reached by the object and the time that the height is reached.

Step-by-Step Solution

Answer

The maximum height is 300 feet, reached after 2.5 seconds.

Step 1: Understand the Problem

We are given a function, \( h(t) = -16t^2 + 80t + 200 \), that represents the height of an object fired upward from a 200-foot tower. We need to find the maximum height and the time at which this maximum height occurs.

Step 2: Identify the Vertex of the Parabola

The function \( h(t) = -16t^2 + 80t + 200 \) is a quadratic equation, which forms a parabola when graphed. Since it has a negative leading coefficient (-16), it opens downward and its vertex represents the maximum point. The formula to find the time \( t \) at which the vertex occurs is \( t = -\frac{b}{2a} \), where \( a = -16 \) and \( b = 80 \).

Step 3: Calculate the Time of Maximum Height

Plug \( a = -16 \) and \( b = 80 \) into the vertex formula: \[ t = -\frac{80}{2(-16)} = \frac{80}{32} = 2.5 \] Thus, the time at which the maximum height is reached is 2.5 seconds.

Step 4: Calculate the Maximum Height

Now, substitute \( t = 2.5 \) back into the height function to find the maximum height:\[ h(2.5) = -16(2.5)^2 + 80(2.5) + 200 \]Calculate each term: - \( -16(2.5)^2 = -16 \times 6.25 = -100 \) - \( 80 \times 2.5 = 200 \)Add these to the initial height (200):\[ h(2.5) = -100 + 200 + 200 = 300 \]The maximum height the object reaches is 300 feet.

Key Concepts

Maximum Height

When an object is projected upwards, it will rise to a certain height before gravity pulls it back down to the ground. This highest point is known as the maximum height. In the context of this problem, the maximum height can be found by analyzing the quadratic equation representing the object's trajectory. The equation is given as \[ h(t) = -16t^2 + 80t + 200 \]The maximum height corresponds to the highest value of \( h(t) \), which occurs at the vertex of the parabola. To find it:

• First, determine the time \( t \) at which the maximum height occurs using the vertex formula.
• Then, substitute this time back into the height function.

In this problem, we calculated that the object reaches its maximum height of 300 feet after 2.5 seconds.

Vertex of a Parabola

The vertex of a parabola represents either its highest or lowest point, depending on the direction it opens. For a quadratic equation in the form \( ax^2 + bx + c \), the vertex provides crucial information about the trajectory's peak. In our exercise, the function is: \[ h(t) = -16t^2 + 80t + 200 \]It opens downward since the leading coefficient \( a \) is negative (-16), meaning the vertex gives us the maximum height.
To find the vertex, use the formula \( t = -\frac{b}{2a} \):

• Substitute \( a = -16 \) and \( b = 80 \) into the formula.
• Calculate the resulting time, which will be the \( t \) value at the vertex.

In our problem, this calculation gives us \( t = 2.5 \) seconds, marking the point in time when the object reaches its peak height.

Kinematic Equations

Kinematic equations describe the motion of objects and are fundamental in analyzing projectile motion. They help us predict an object's position and velocity over time under constant acceleration conditions like gravity. The key kinematic equation used in this problem is: \[ h(t) = -16t^2 + 80t + 200 \]Here, -16 represents the gravitational acceleration (in feet per second squared), the \(-16t^2\) term accounts for acceleration, 80 \(t\) represents initial velocity, and 200 is the initial height.

• These components together build a picture of how the object's flight path develops over time.
• The negative sign in the \(-16t^2\) term shows gravity's role in pulling the object back down.

Understanding kinematic equations allows us to predict when and where an object will be at any given point during its flight.

Parabolic Trajectory

A parabolic trajectory refers to the path followed by an object under the influence of uniform gravity, assuming air resistance is negligible. This kind of motion arises from the symmetry and properties of quadratic equations. When graphed, these equations form parabolas.
For the function \( h(t) = -16t^2 + 80t + 200 \):

• The object's launch velocity and initial position shape the equation, producing a curve that is initially upward until gravity reverses the motion.
• The trajectory is symmetrical around the vertex, where the object reaches maximum height.

Analyzing the properties of a quadratic equation lets us infer key points along the object's trajectory, such as maximum height and time of flight.