Problem 47 - Essential Calculus Early Transcendentals

Question

A stone was dropped off a cliff and hit the ground with a speed of 120 \(\mathrm{ft} / \mathrm{s} .\) What is the height of the cliff?

Step-by-Step Solution

Answer

The height of the cliff is 225 feet.

Step 1: Understand the Problem

We need to determine the height of a cliff from which a stone was dropped. We know the stone hit the ground with a speed of 120 \(\text{ft/s}\). We will use the principles of free fall and gravitational acceleration to find the height of the cliff.

Step 2: Use the Free Fall Formula

The formula for the velocity of an object in free fall is given by \( v = gt \), where \( v \) is the final velocity, \( g \) is the acceleration due to gravity (approximately \(32 \text{ft/s}^2\) on Earth), and \( t \) is the time of fall. We will use this to solve for \( t \).

Step 3: Calculate the Time of Fall

Given \( v = 120 \text{ft/s} \) and \( g = 32 \text{ft/s}^2 \), solve for \( t \): \[ t = \frac{v}{g} = \frac{120}{32} \approx 3.75 \text{s} \]. This is the time it took for the stone to fall.

Step 4: Find the Height Using the Time

The height \( h \) from which the stone was dropped can be found using the formula \( h = \frac{1}{2}gt^2 \). Plug in the values: \[ h = \frac{1}{2} \times 32 \times (3.75)^2 \].

Step 5: Perform the Final Calculation

Calculate the height: \[ h = 16 \times 3.75^2 = 16 \times 14.0625 = 225 \]. Therefore, the height of the cliff is 225 feet.

Key Concepts

Velocity in Free Fall

When an object is in free fall, it means that the only force acting on it is gravity. This type of motion is crucial in understanding basic physics principles. In the given problem, a stone is dropped from a cliff, and by the time it reaches the ground, it achieves a certain speed. This speed is known as the "velocity in free fall."

• Velocity is simply speed with a direction. For objects in free fall, like our stone, the direction is downwards due to gravity.
• The velocity an object reaches just before impact in free fall can be calculated using the formula: \( v = gt \).
• Here, \( v \) is the final velocity (120 \( \text{ft/s} \) in this problem), and \( g \) is the acceleration due to gravity.

In our exercise, the stone reaches a final velocity of 120 \( \text{ft/s} \) just before it hits the ground, which gives us a crucial piece of information to calculate other aspects of its motion, like the time it took to fall.

Acceleration Due to Gravity

Acceleration due to gravity is a powerful and continuously acting force that affects all objects in free fall. On Earth, this acceleration is typically around 32 \( \text{ft/s}^2 \), although it can vary slightly depending on location.

• This acceleration is constant, meaning it doesn’t change as the object falls.
• In our problem, the acceleration due to gravity enables us to calculate both the time it takes for the object to reach the ground and the height of the cliff, which are interconnected.

The formula \( v = gt \) relates the acceleration due to gravity to the velocity achieved by an object in free fall. It’s essential for determining the time of fall, which further leads us to calculate the height of the fall.

Kinematic Equations

Kinematic equations are the formulas used to solve problems involving motion, like free fall. They help link different quantities such as time, velocity, and distance.

• In the cliff problem, we start with the formula for time \( t = \frac{v}{g} \), derived from \( v = gt \). This helps us calculate the time the stone took to fall.
• Once time is known, another kinematic formula to find the height, \( h = \frac{1}{2}gt^2 \), is applied.
• This formula is essential because it ties together the constant acceleration due to gravity, the time the stone was in the air, and the resulting height from which it fell.

These equations provide a comprehensive framework for understanding motion, especially for free-falling objects. By solving one variable, like time, you can easily navigate to determine other variables, such as height, making these equations invaluable in physics.