Problem 78 - Algebra 1

Question

Use a vertical motion model to find how long it will take for the object to reach the ground. An acorn falls 45 feet from the top of a tree.

Step-by-Step Solution

Answer

To find the time the acorn takes to fall, we use the formula \(d = 0.5gt^2\) which gives us \(t = sqrt(45 / (0.5*32.2))\). Once we calculate that, we get the time approximately equals to 1.68 seconds.

Step 1: Understand the physics behind the problem

Understand that an object in free fall will experience acceleration due to gravity. The approximate value of this is \(9.8 m/s^2\). However, this is in metric measurements. As our problem has measurements in feet, we should use the conversion \(32.2 ft/s^2\) for gravity acceleration.

Step 2: Write down the formula for motion under gravity

The formula linking distance fallen \(d\), gravity \(g\), and time \(t\) for an object in free fall from rest is \(d = 0.5gt^2\). In this case, \(d = 45ft\) and \(g = 32.2 ft/s^2\) and we need to solve for \(t\).

Step 3: Substitute values into the formula

Substitute \(d = 45ft\) and \(g = 32.2 ft/s^2\) into the equation to get \(45 = 0.5*32.2*t^2\).

Step 4: Solve for time

To isolate \(t^2\), we divide both sides of the equation by \(0.5*32.2\) to get \(t^2 = 45 / (0.5*32.2)\). We then take the square root of both sides to solve for \(t\) which gives us \(t = sqrt(45 / (0.5*32.2))\).

Key Concepts

Free Fall Acceleration

When an object is dropped, gravity causes it to accelerate towards the ground. This acceleration is constant and for Earth's gravity, it is approximately ​9.8 meters per second squared (m/s²) in metric units. In problems involving vertical motion in a region where air resistance is negligible, we use this standard value to represent the acceleration due to gravity, symbolized by the letter ​g.

For those more familiar with the Imperial system, such as in the United States, this value is converted to about ​32.2 feet per second squared (ft/s²) because the problem provided to us is using feet as the unit of measurement. It's crucial to use the correct value of ​g​ in our calculations, as mixing units can lead to incorrect results.

Gravity in Algebraic Problems

Incorporating gravity into algebraic problems involves understanding and applying formulas that govern motion under the influence of Earth's gravitational pull. The key equation for objects in free fall from rest is ​d = 0.5 * g * t², where ​d​ represents the distance fallen, ​g​ is the acceleration due to gravity, and ​t​ is the time taken to fall that distance.

Understanding how to manipulate this formula is essential. When we come across a problem, we need to identify known quantities and if necessary, translate them into appropriate units. We then substitute these known values into the formula, and solve for the unknown quantity, which in many physics problems, is the time ​t.

Solving for Time in Physics Equations

Solving for time within physics equations is a frequent requirement, and mastering it requires some algebraic manipulation. With the vertical motion formula ​d = 0.5 * g * t², our goal is to solve for ​t​. It's important, to begin with rearranging the formula to isolate ​t²​ on one side of the equation. For the given problem, we would divide both sides by ​0.5 * g​ to get ​t² = d / (0.5 * g)​. Finally, taking the square root of both sides will give us ​t​, which is the time in seconds it takes for the object to reach the ground.

This process is not unique to free-fall problems—it is a typical method used to find time in various uniform acceleration scenarios. Remember to check the units before inserting your values, as time solved in seconds is the standard in physics.