Problem 91 - Algebra and Trigonometry

Question

Falling Ball Using calculus, it can be shown that if a ball is thrown upward with an initial velocity of 16 \(\mathrm{ft} / \mathrm{s}\) from the top of a building 128 \(\mathrm{ft}\) high, then its height \(h\) above the ground \(t\) seconds later will be $$ h=128+16 t-16 t^{2} $$ During what time interval will the ball be at least 32 \(\mathrm{ft}\) above the ground?

Step-by-Step Solution

Answer

The ball is at least 32 ft above the ground during the interval \( 0 \leq t \leq 3 \).

Step 1: Understand the Problem

We need to find the time interval during which the height of the ball is at least 32 feet. This means solving for \( t \) in the inequality \( h \geq 32 \).

Step 2: Set Up the Inequality

Replace \( h \) in the equation with 32, so that the inequality becomes: \( 128 + 16t - 16t^2 \geq 32 \).

Step 3: Rearrange the Inequality

Subtract 32 from both sides of the inequality to simplify it:\( 96 + 16t - 16t^2 \geq 0 \).

Step 4: Simplify and Set Up Quadratic Equation

Rearrange terms into a standard quadratic form: \( -16t^2 + 16t + 96 \geq 0 \). Multiply through by -1 to make it easier to solve: \( 16t^2 - 16t - 96 \leq 0 \). Factor out 16: \( t^2 - t - 6 \leq 0 \).

Step 5: Factor the Quadratic

Factor the quadratic equation: \( t^2 - t - 6 = (t - 3)(t + 2) \). The inequality now becomes: \( (t - 3)(t + 2) \leq 0 \).

Step 6: Find the Critical Points

Set each factor equal to zero to find the critical points:- \( t - 3 = 0 \) gives \( t = 3 \).- \( t + 2 = 0 \) gives \( t = -2 \).

Step 7: Test Intervals

Test intervals determined by critical points:1. Test \( t < -2 \) (choose \( t = -3 \)): The expression \((t - 3)(t + 2)\) is positive.2. Test \( -2 < t < 3 \) (choose \( t = 0 \)): The expression \((t - 3)(t + 2)\) is negative.3. Test \( t > 3 \) (choose \( t = 4 \)): The expression \((t - 3)(t + 2)\) is positive.

Step 8: Determine the Valid Interval

From Step 7, the inequality \((t - 3)(t + 2) \leq 0\) is satisfied for \(-2 < t < 3 \). However, since \( t \) represents time, \( t < 0 \) isn't physically meaningful, so the valid time interval is \( 0 \leq t \leq 3 \).

Key Concepts

Calculus

Calculus is a branch of mathematics that helps us understand changes. It's all about how things move and grow. When dealing with physical problems, like a ball flying in the air, calculus helps us calculate how height, speed, or position change over time. These calculations often involve derivative and integral concepts.

By using calculus, we can derive equations that model real-world situations. In the exercise with the falling ball, calculus is employed to derive the height equation, detailing how the height changes with time. This understanding is essential to solve the problem and determine when the ball is at least 32 feet high.

Velocity

Velocity refers to the speed and direction of an object's movement. It tells us how fast something is going and in which direction. In this exercise, the ball is thrown upward with an initial velocity of 16 feet per second.

The velocity affects how the height of the ball changes over time. Initially, the ball rises because of the throw, slowing down due to gravity until it reaches a peak, then it begins to fall. Understanding the role of initial velocity helps us appreciate how far or high an object will travel after being set in motion.

• Initial velocity: 16 ft/s upward
• Impact of gravity: Slows upward motion, then accelerates downward

Quadratic Equations

Quadratic equations are vital in solving the exercise because the height of the ball over time follows a quadratic pattern. A quadratic equation typically looks like this: \( ax^2 + bx + c = 0 \). It has a special U-shape when plotted on a graph, called a parabola.

In the exercise, the height of the ball over time is given by the quadratic equation: \( h = 128 + 16t - 16t^2 \). Solving quadratic inequalities, like the one where we need the ball's height to be at least 32 feet, involves finding the range of \( t \) values that satisfy the inequality.

Factoring

Factoring is a method used to solve quadratic equations and inequalities by breaking them down into simpler expressions. In this exercise, we have the equation \( t^2 - t - 6 = 0 \).

When an equation or expression is factored, it's expressed as a product of its simplest parts. Factoring the above gives us \( (t - 3)(t + 2) = 0 \). These factors help identify the points where the ball's height changes critical behavior—specifically, the times when it equals 32 feet. Factoring is key because it simplifies solving inequalities, enabling us to find where the quadratic expression is less than or equal to zero.

• Critical points determine transition points in the parabola
• Factoring yields roots, critical for interpreting the solution