Problem 30 - Algebra A Combined Function

Question

Solve. If Rheam Gaspar throws a ball upward with an initial speed of 32 feet per second, then its height \(h\) in feet after \(t\) seconds is given by the function \(h(t)=-16 t^{2}+32 t\). Find the maximum height of the ball.

Step-by-Step Solution

Answer

The maximum height of the ball is 16 feet.

Step 1: Understand the Function Form

The height function given is quadratic, written as \(h(t) = -16t^2 + 32t\). This is a downward-opening parabola because the coefficient of \(t^2\) is negative.

Step 2: Identify the Maximum Point

For a quadratic function in the form \(ax^2 + bx + c\), the maximum (or minimum) occurs at \(x = -\frac{b}{2a}\) when \(a < 0\). Thus, \(t = -\frac{32}{2(-16)}\).

Step 3: Calculate Time of Maximum Height

Substitute the values, \(a = -16\) and \(b = 32\), into the formula to find the time the ball reaches its maximum height: \(t = -\frac{32}{2(-16)} = 1\) second.

Step 4: Substitute to Find Maximum Height

Plug \(t = 1\) back into the height function to find the maximum height: \(h(1) = -16(1)^2 + 32(1) = -16 + 32 = 16\).

Step 5: Conclusion

The maximum height reached by the ball is 16 feet.

Key Concepts

Maximum Height

The maximum height of a ball thrown into the air is the highest point that the ball reaches before it starts to descend back to the ground. When dealing with balls or other objects thrown upwards, the concept of maximum height is essential to understand.

• In our scenario, the height function is quadratic. It tells us how the height of the ball changes over time.
• The maximum height occurs at the peak of the parabola, which is the topmost point of the curve.

To determine this height, we need to find out when the upward velocity becomes zero, meaning the ball has stopped rising and is about to fall back down. This is where our understanding of maximum height intersects with the concept of "parabolic motion," offering a comprehensive look at how objects move when launched upwards.

Parabolic Motion

Parabolic motion refers to the path of an object that is projected into the air. Usually, this results in a curved trajectory that looks like an arch. The path taken is called a parabola, characterized by a consistent symmetric shape.

• In our specific problem, the ball's motion creates a downward-opening parabola.
• Its equation, \(-16t^2 + 32t\), describes this curve. Here, the negative coefficient of \(t^2\) shows that the parabola opens downwards.

This type of motion is common in situations involving the force of gravity, which consistently pulls the object back towards the earth, causing it to rise at first, slow down, and eventually fall.

Vertex Formula

The vertex formula is crucial for finding the maximum or minimum point of a quadratic function. This formula, \(x = -\frac{b}{2a}\), finds the x-coordinate of the vertex of the parabola.

• The vertex represents the highest or lowest point of the parabola.
• For a quadratic function in standard form \(ax^2 + bx + c\), the vertex formula helps in easily calculating the vertex.

In our problem, by substituting \(a = -16\) and \(b = 32\) into this formula, we determine the time \(t\) when the ball reaches its maximum height. Recognizing this use of the vertex formula in equations of motion is essential when working with quadratics.

Quadratic Equation

A quadratic equation is a second-degree polynomial and typically takes the form \(ax^2 + bx + c = 0\). In many mathematical and physical contexts, solving these equations is important.

• In our exercise, the function \(h(t) = -16t^2 + 32t\) traces the height of the ball over time.
• The quadratic component, represented by \(t^2\), describes the way the object's speed and direction change.

Understanding the nature of quadratic equations can greatly aid in interpreting scenarios involving parabolic motion, such as projectiles or thrown objects. In this particular case, it lets us take advantage of the vertex formula to see exactly when and where our ball reaches its highest point.