Problem 97 - Algebra 1

Question

Use the following information. Scientists simulate a gravity-free environment called microgravity in free- fall situations. A similar microgravity environment can be felt on free-fall rides at amusement parks or when stepping off a high diving platform. The distance \(d\) (in meters) that an object that is dropped falls in \(t\) seconds can be modeled by the equation \(d=\frac{1}{2} g\left(t^{2}\right),\) where \(g\) is the acceleration due to gravity (9.8 meters per second per second). The NASA Lewis Research Center has two microgravity facilities. One provides a 132 -meter drop into a hole and the other provides a 24 -meter drop inside a tower. How long will each free-fall period be?

Step-by-Step Solution

Answer

The time taken for the object to free-fall in 132-meter and 24-meter distance are approximately 5.18 seconds and 2.21 seconds respectively.

Step 1: Understand the equation and what it models

Firstly, here is the equation being used to model the scenario: \(d=\frac{1}{2} g\left(t^{2}\right)\). This equation represents the distance \(d\) that an object falls under gravity \(g\) in a specified time \(t\). It basically says that the distance fallen is proportional to the square of the time of falling.

Step 2: Insert values into the equation for the first scenario

Now, for the first case when the object is dropped into a 132-meter hole, the equation becomes \(132=\frac{1}{2} * 9.8\left(t^{2}\right)\). The next step is to solve for \(t\) from this equation.

Step 3: Solve for 't' in the first case

Solving for \(t\) from the above equation, first divide both sides by \(\frac{1}{2} * 9.8\) to get \(t^{2}\). Then take the square root of both sides to solve for \(t\).

Step 4: Repeat step 2 and 3 for the second scenario

For the second scenario where the object is dropping in a 24-meter tower, the equation becomes \(24=\frac{1}{2} * 9.8\left(t^{2}\right)\). Solve for \(t\) similar to step 3.

Key Concepts

Quadratic Equations in Physics

A quadratic equation in physics is fundamental for modeling situations where acceleration is constant, such as free-fall motion. It consists of terms up to the second power of the unknown variable and can be recognized by its standard form, which is \( ax^2 + bx + c = 0 \). In free-fall scenarios, the equation \( d = \frac{1}{2} g t^2 \) directly relates the distance \( d \) an object falls to the square of the time \( t \) it's been falling, introducing \( g \) as the acceleration due to gravity.

In the context of our exercise, we treat the equation \( d = \frac{1}{2} g t^2 \) as a quadratic equation with \( a = \frac{1}{2} g \) and \( c = -d \) (distance fallen), where \( b = 0 \) because the equation does not have a first power term. When solving a quadratic equation in physics, you often use algebraic techniques to rearrange the terms and isolate the variable of interest—time (\( t \) in our case). The square root method becomes the final step to determine the value for \( t \) when the equation is already in the form \( at^2 = d \) after initial manipulations.

Acceleration Due to Gravity

The acceleration due to gravity, denoted as \( g \), is a constant that represents the gravitational force exerted by a massive body, like Earth, on objects. The standard value of \( g \) on Earth's surface is \( 9.8 \text{m/s}^2 \). This means that, in the absence of air resistance, an object's speed increases by 9.8 meters per second each second it is in free fall.

In many physics problems, including those involving free-fall motion like the textbook exercise, the value of \( g \) is crucial to accurately predict the outcomes. It directly influences the distance an object will fall over time, as seen in the equation \( d = \frac{1}{2} g t^2 \). Realizing that \( g \) is a constant value allows for simplified calculations and better understanding of motion under the influence of Earth's gravity. While \( g \) can slightly vary depending on altitude and geographical location, for most educational problems, the standard value is used to ensure the simplicity and consistency of the problem-solving process.

Solving for Time in Physics Problems

In physics problems, particularly those involving motion equations, solving for time \( t \) is a common task. Time is typically what you're trying to find out to understand the duration of an event, like the free-fall of an object. To do this, you rearrange the motion equation to isolate and solve for \( t \) as a function of the other quantities. The problem provided involves a quadratic equation, which you can manipulate to solve for time by isolating \( t^2 \) and then extracting the square root.

Method for Solving for Time

Applied to our exercise, the steps to solve for time in a free-fall motion scenario are:

• Start with the equation \( d = \frac{1}{2} g t^2 \).
• Rearrange to solve for \( t^2 \) by multiplying both sides by \( 2/g \) resulting in \( t^2 = (2d)/g \).
• Take the square root of both sides to find \( t \), which gives \( t = \sqrt{(2d)/g} \).

By following these steps, you ensure that you systematically approach the problem and derive the time taken for an object to fall a certain distance under the influence of gravity.