Problem 34 - Calculus Single Variable

Question

The acceleration due to gravity near the surface of the moon is about \(-5.3064 \mathrm{ft} / \mathrm{s}^{2} .\) If an object is dropped from the top of a cliff on the moon, and takes 5 seconds to strike the surface of the moon, then how high is the cliff?

Step-by-Step Solution

Answer

The height of the cliff is approximately 66.33 ft.

Step 1: Understand the relevant formula

When an object is dropped from rest, its distance fallen over time can be calculated using the formula: \( s = \frac{1}{2}gt^2 \) where \( s \) is the distance (height of the cliff in this case), \( g \) is the acceleration due to gravity, and \( t \) is the time the object has fallen.

Step 2: Substitute known values into the formula

From the exercise, \( g = -5.3064 \text{ ft/s}^2 \) (note the negative sign as it's technically downward), and \( t = 5 \text{ s} \). Substituting these values into the formula gives: \( s = \frac{1}{2}(-5.3064)(5^2) \).

Step 3: Calculate the squared time value

Calculate \( 5^2 \), which is \( 25 \). This results in: \( s = \frac{1}{2}(-5.3064)(25) \).

Step 4: Multiply and simplify

First, multiply \( -5.3064 \times 25 = -132.66 \). Then, find \( s = \frac{1}{2}(-132.66) = -66.33 \).

Step 5: Understand the physical meaning

Since \( s \) represents distance, we consider its absolute value. Thus, the height of the cliff is \( 66.33 \text{ ft} \).

Key Concepts

Acceleration due to Gravity

Gravity is a force that pulls objects towards each other. On Earth, this force pulls objects downwards, and we call it gravitational acceleration. On different celestial bodies like the moon, gravity acts similarly but with different strength. For instance, at the moon's surface, gravity accelerates objects towards it at about \(-5.3064 \, \text{ft/s}^2\). This means that any free-falling object near the moon's surface will accelerate downwards at this rate.

Since these values often involve negative numbers because of the direction involved in falling, it's important to note that gravity works downward, hence the negative sign. When working with direction in physics, numbers can tell us the direction in which the force is acting. Understanding this helps us correctly calculate motion phenomena, such as free fall.

Free Fall

When an object is in free fall, it means that the only force acting upon it is gravity. There's no air resistance on the moon, as there's no atmosphere, leaving gravity as the sole energy moving the object towards the lunar surface. In this case, the object dropped from rest starts accelerating at the same rate as the gravitational pull, which on the moon is \(-5.3064 \, \text{ft/s}^2\).

For instance, if you were to drop a rock from the cliff on the moon, the rock would be in free fall. This means it speeds up as it moves downwards, following this gravitational acceleration. Free fall makes calculations more straightforward, as the acceleration is constant and unopposed by other forces.

This principle is important when calculating how far or how fast an object will travel while under the influence of gravity alone.

Distance Calculation

Calculating the distance an object falls involves understanding the relationship between uniformly accelerated motion and time. If you know the time an object was in free fall and the rate of acceleration due to gravity, you can compute the distance fallen using the equation \( s = \frac{1}{2}gt^2 \).

• \( s \) represents the distance the object travels, which could be the height from which it was dropped.
• \( g \) is the gravitational acceleration – negative if we take downward as negative.
• \( t \) is the time duration of the fall.

In our example from the moon, after substituting the known variables \( g = -5.3064 \, \text{ft/s}^2 \) and \( t = 5 \, \text{s} \), we perform the calculation to get \(-66.33\, \text{ft}\). Despite the negative sign indicating direction, the distance or height of the cliff is \(66.33\, \text{ft}\). It shows how far the object has traveled in this timeframe. Understanding how to compute this provides a clear insight into the physics behind motion under gravity and solidifies our grasp on the general principles of kinematics.