Problem 95 - Algebra and Trigonometry

Question

Falling-Body Problems Suppose an object is dropped from a height \(h_{0}\) above the ground. Then its height after \(t\) seconds is given by \(h=-16 t^{2}+h_{0}\) , where \(h\) is measured in feet. Use this information to solve the problem. If a ball is dropped from 288 \(\mathrm{ft}\) above the ground, how long does it take to reach ground level?

Step-by-Step Solution

Answer

The ball takes approximately 4.24 seconds to reach the ground.

Step 1: Identify Known Values

We know that the initial height \(h_{0}\) is 288 feet, and the height \(h\) when the ball reaches the ground is 0 feet. The equation we use is \(h = -16t^2 + h_0\).

Step 2: Set the Height to Zero

Since we want to find the time \(t\) when the ball reaches the ground, we set the height \(h\) to 0 in the equation \(0 = -16t^2 + h_0\).

Step 3: Substitute Known Values into Equation

Substitute \(h_0 = 288\) into the equation: \(0 = -16t^2 + 288\).

Step 4: Solve for Time \(t\)

First, rearrange the equation to isolate \(t^2\): \(16t^2 = 288\). Divide both sides by 16 to get \(t^2 = 18\).

Step 5: Take the Square Root

Take the square root of both sides to solve for \(t\): \(t = \sqrt{18}\). Simplify this to \(t = 3\sqrt{2}\), which can be approximately calculated as \(t \approx 4.24\).

Key Concepts

Quadratic Equations

At the core of solving falling body problems, such as determining the time it takes for a ball to reach ground level, is the quadratic equation. Quadratic equations are mathematical expressions that can be represented in the form
\[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. These equations take a U-shaped curve on a graph known as a parabola.
For the falling ball problem, the height equation \( h = -16t^2 + h_0 \) is a quadratic equation in terms of \( t \), where \( h_0 \) is the initial height, and the coefficient \(-16\) represents the acceleration due to gravity measured in feet per second squared. When solving the equation, you set \( h = 0 \) to find the time when the object hits the ground, turning the equation into \( 0 = -16t^2 + 288 \).
The solution involves rearranging the equation to isolate \( t^2 \), dividing by the coefficient of \( t^2 \), and finally taking the square root to solve for \( t \). Understanding the steps of manipulating the equation helps unravel the behavior of objects in free fall, a topic in kinematics.

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion. It helps us understand how objects move through space over time. In the case of a falling ball, kinematics provides a framework to model the object's changing position as time passes.
The height equation \( h = -16t^2 + h_0 \) accurately represents the ball's height as a function of time. Here, the term \(-16t^2\) is derived from the kinematic equation for objects under constant acceleration. This negative term accounts for the downward pull of gravity, which is approximately \(32 \text{ ft/s}^2\) in the real world, but the factor of \(-16\) in the formula comes from half of that constant (since the standard form involves \(\frac{1}{2}gt^2\)).
Kinematics thus simplifies complex motions into simple mathematical models. Key quantities such as initial velocity, time, final position, and gravity can be plugged into kinematic equations to predict future motion states like the height of an object in freefall.

Physics Calculations

Physics calculations for falling objects often meld mathematical analysis with physical principles. In the example of a ball falling 288 feet, physics calculations dictate how we set up and solve the problem using formulas that represent real-world phenomena.
Here are the steps involved:

• Write down the known values: initial height \( h_0 = 288 \) feet, and final height \( h = 0 \) feet when the ball reaches the ground.
• Use the height equation \( h = -16t^2 + h_0 \) to set up the problem by substituting known values \(0 = -16t^2 + 288\).
• Solve algebraically for \( t \) by isolating \( t^2 \), then taking the square root to find \( t = \sqrt{18} = 3\sqrt{2} \approx 4.24 \) seconds.

Such calculations demand precision, with attention to unit measurements, signs (positive and negative), and correct arithmetic operations. Using a calculator might help verify the square root calculations and approximations. Through applied physics calculations, we convert theoretical principles into tangible predictions that can be measured and observed.