Problem 83 - 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 feet above the ground during the time interval [0, 3] seconds.

Step 1: Understand the Problem

We need to find the time interval during which the ball's height is at least 32 feet above the ground. The height equation provided is \( h = 128 + 16t - 16t^2 \). We want \( h \) to be greater than or equal to 32 feet, i.e., \( 128 + 16t - 16t^2 \geq 32 \).

Step 2: Set Up the Inequality

To find the time interval, set up the inequality: \( 128 + 16t - 16t^2 \geq 32 \). Simplify this to: \( 96 + 16t - 16t^2 \geq 0 \).

Step 3: Rearrange the Inequality

Rearrange the inequality to standard quadratic form: \(-16t^2 + 16t + 96 \geq 0 \). It can be rewritten as: \(-16(t^2 - t - 6) \geq 0 \).

Step 4: Solve the Quadratic Equation

Factor the quadratic equation inside the parentheses: \(t^2 - t - 6 = 0 \). Using the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 1\), \(b = -1\), and \(c = -6\).

Step 5: Simplify and Calculate the Roots

Calculate the discriminant: \((-1)^2 - 4(1)(-6) = 1 + 24 = 25\). Find the roots using the quadratic formula: \(t = \frac{1 \pm \sqrt{25}}{2} = \frac{1 \pm 5}{2}\). The solutions are \(t = 3\) and \(t = -2\).

Step 6: Determine the Relevant Time Interval

Since time cannot be negative, we discard \(t = -2\). The relevant interval starts from the positive root. We analyze \(-16(t - 3)(t + 2) \geq 0\) for \(t\) values between where the factors change sign to find the interval. The ball reaches at least 32 feet between \( t = 0 \) and \( t = 3 \).

Step 7: Confirm and Conclude the Interval

Check if the factors \( (t - 3)(t + 2) \) are non-negative. Possible intervals are from the zeroes and points in between. Verify that \(t \in [0, 3]\) holds the inequality true based on calculations from roots and leading coefficients.

Key Concepts

Calculus Applications

Calculus is an essential tool for understanding and solving complex problems involving motion, such as the movement of a projectile like a thrown ball. By using calculus, we can derive formulas that describe how an object's position changes over time under the influence of gravity and other forces. In this exercise, we calculate the height of a ball at any time after it's thrown upward from a building. The equation given, \[ h = 128 + 16t - 16t^2 \] tells us how high the ball is at any given second \( t \) after being thrown. The coefficients in this equation stem from applying physics concepts, mainly considering the ball's initial velocity and gravitational acceleration.

• Initial velocity is the speed at which the ball is thrown upward, here given as 16 ft/s.
• Gravitational acceleration is the force pulling the ball back down, typically \( -32 \text{ ft/s}^2 \) for downward motion, hence the \(-16t^2\) due to calculus integration.

These applications show how calculus can transform the understanding of physical motion into mathematical expressions, providing a way to solve real-world motion scenarios.

Projectile Motion

Understanding projectile motion is crucial in physics and has direct applications in every day and advanced studies. A projectile is any object thrown into space, influenced only by its initial velocity and the gravitational pull affecting it. In this case, our projectile is the ball thrown from the building, impacted by gravity. Two main characteristics of projectile motion are:

• Horizontal motion: It is constant because no horizontal forces like air resistance are considered here.
• Vertical motion: It changes over time due to the effect of gravity, pulling the object down to earth at an accelerating rate.

During the ball's flight, it first rises, temporarily stops at its peak height, and then descends back down. The quadratic equation illustrates this journey by depicting the ball's height over time. By solving quadratic inequalities, as done in the solution, we determine the time intervals when the ball is at specific vertical positions, such as being at least 32 feet above the ground.

Quadratic Equations

At the heart of solving this exercise is understanding quadratic equations, which are polynomial equations of the second degree. The general form of a quadratic equation is \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants.In this context, we are not merely solving for when the equation equals zero, but rather when it holds a certain inequality:\[ 128 + 16t - 16t^2 \geq 32 \]To solve, rearrange and simplify the inequality to form a quadratic equation:\[-16(t^2 - t - 6) \geq 0\]By solving \( t^2 - t - 6 = 0 \) with the quadratic formula, we find the critical points where the inequality changes from true to false or vice versa:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The provided roots, \( t = 3 \) and \( t = -2 \), are analyzed by checking the sign changes in intervals determined by these points. Discarding negative time, which is not possible in real situations, helps to identify the valid interval \([0, 3]\) where the ball remains above 32 feet. This approach not only solves for the critical moments but gives a complete view of the ball's trajectory relative to a specified condition.