Problem 60 - Algebra 1: Concepts and Skills

Question

A boulder falls off the top of an overhanging cliff during a storm. The cliff is 96 feet high. Find how long it will take for the boulder to hit the road below. Solve the falling object model for \(h=0 .\) Round to the nearest tenth.

Step-by-Step Solution

Answer

The boulder will hit the road below in approximately 2.4 seconds.

Step 1: Identifying the known variables

From the problem, we know that the initial velocity \(v_0 = 0 ft/s\), the height \(h = 96 feet\), and the acceleration due to gravity \(g= 32 ft/s^2 \).

Step 2: Substitute into the equation

Substituting the known quantities into the falling object model equation yields: \(96 = 0 * t + 0.5 * 32 * t^2\). With \(v_0 = 0 ft/s\), the equation simplifies to \(96 = 16 * t^2\).

Step 3: Solve for time

To find the time it takes for the boulder to hit the road, we solve the equation for \(t\). Dividing both sides by 16 gives \(t^2 = 6\). Taking square root of both sides, we get \(t = \sqrt{6}\). Since time cannot be negative, we only consider the positive root.

Step 4: Round to the nearest tenth

We round the value of \(t\) to the nearest tenth. This gives \(t \approx 2.4 seconds\).

Key Concepts

Falling Object Model

The falling object model is an equation used in physics to describe the motion of an object that is dropped from a height with no initial velocity. This model is particularly useful when analyzing situations where you need to understand how long it takes for the object to reach the ground.
The standard equation for an object falling from rest under the influence of gravity is:

• \[ h = v_0 t + \frac{1}{2} gt^2 \]

where \( h \) represents the height from which the object falls, \( v_0 \) is the initial velocity, \( t \) is the time in seconds, and \( g \) is the acceleration due to gravity.
This equation assumes no air resistance, perfect conditions, which allows us to purely focus on gravitational effects. In many physics problems, such as the one involving the boulder, the initial velocity \( v_0 \) is often \( 0 \), as the object is usually dropped rather than thrown.

Acceleration Due to Gravity

The constant \( g \), known as the acceleration due to gravity, is one of the most significant constants in physics when studying motion. This value represents the acceleration that any object experiences when falling freely towards the Earth without any other forces acting on it except gravity itself.
On Earth, \( g \) is approximately \( 32 \, \text{ft/s}^2 \). This means any object, irrespective of its mass, will increase its velocity by \( 32 \, \text{ft/s} \) every second it remains in free fall.
It's essential to understand that the acceleration due to gravity varies depending on the location (e.g., smaller on the Moon). However, for standard physics problems and introductory courses, this value is treated as a constant for all calculations. In any vertical motion problem, knowing this constant allows us to simplify and solve equations efficiently.

Solving Quadratic Equations

Solving quadratic equations is a fundamental part of mathematics and physics. In the context of our problem, we end up with a quadratic equation while using the falling object model:

• \[ 96 = 16t^2 \]

This is a simplified quadratic equation where the equation:

• \[ at^2 + bt + c = 0 \]

is transformed to the form \( 16t^2 = 96 \). To solve for \( t \), we divide both sides by 16, which simplifies to \( t^2 = 6 \). Then, by taking the square root of both sides, we find \( t = \sqrt{6} \). In our example, only the positive root is meaningful since time cannot be negative.
Knowing how to solve this type of equation is crucial in determining the time it takes for objects to reach certain points, such as when they hit the ground. In many physics-related problems, this step essentially allows us to make accurate predictions about motion in a gravitational field.