Problem 69 - Pearson Physics

Question

When you drop a rock into a well, you hear the splash \(1.5\) s later. (a) If the distance to the water in the well were doubled, would the time required to hear the splash be greater than, less than, or equal to \(3.0 \mathrm{~s}\) ? Explain. (b) How far down was the water originally?

Step-by-Step Solution

Answer

Doubling the distance results in time greater than 3.0s; initial water distance is approximately 10.8 meters.

Step 1: Understand the Problem

A rock is dropped into a well and you hear a splash after 1.5 seconds. We need to determine if doubling the distance would change the time to more, less, or exactly 3.0 seconds, and also find the original distance to the water.

Step 2: Identify Key Concepts

To solve this, we will use the equations of motion and speed of sound. The rock is initially at rest and falls under gravity. Also, consider that sound travels back up to your ear after the splash.

Step 3: Set up Equations

Define the distance to water as \(d\). The time taken for the rock to fall and the time for sound to return sum up to the total time. First calculate the fall time \(t_1\) and use sound speed \(s = 343 \, \text{m/s}\) to find the sound time \(t_2\). Therefore, \(t_1 + t_2 = 1.5 \, \text{s}\).

Step 4: Calculate Fall Time (\(t_1\))

From the rock dropping equation, \(d = \frac{1}{2}gt_1^2\) with \(g = 9.8 \, \text{m/s}^2\). This equation gives how long it takes the rock to hit the water.

Step 5: Setup and Solve Combined Equation

Use \(t_1 + \frac{d}{s} = 1.5\). Substitute \(d\) from the fall equation into this, solving for \(d\).

Step 6: Effect of Doubling Distance (a)

If distance \(d\) is doubled, solve \(t_1' + \frac{2d}{s}\) with new \(t_1'\). Since solutions involve squaring \(t_1\), doubling roughly increases \(t\) in non-linear fashion greater than doubling.

Step 7: Calculate Original Depth (b)

Solve the equation for \(d\) after finding \(t_1\) using values \(\frac{1}{2}gt_1^2 = \frac{s(t_1 + \frac{d}{s})}{1.5} = d\) to determine precise depth.

Key Concepts

Equations of Motion

Equations of motion are essential tools in kinematics, describing the behavior of a body in motion. When solving problems like a free-fall situation, as in the exercise, these equations help us determine unknown parameters such as time, distance, and velocity. The key equation used here is:

• For a freely falling object, the distance (\(d\)) it travels over time (\(t\)) can be expressed as: \[d = \frac{1}{2} g t^2\]where \(g = 9.8 \, \text{m/s}^2\), the acceleration due to gravity.

This formula is derived from the basic principles of acceleration under gravity. In the exercise, the rock is initially at rest, meaning its initial velocity is zero, which simplifies calculations.
By setting up the equations correctly, we distinguish between the time taken for the rock to fall and the time it takes for the sound to travel back to the observer. Each part of the motion has its time component, which together add up to the total time observed.

Speed of Sound

The speed of sound is important in this problem because it dictates how long the sound from the splash takes to reach back to the observer after the rock hits the water. Sound travels through air at approximately \(343 \, \text{m/s}\). This speed can vary slightly depending on factors such as temperature and air pressure, but for most basic physics calculations, this is a standard value.
The time it takes for the sound to travel a certain distance (\(d\)) back to the top of the well is calculated as:

• Time for sound = \(\frac{d}{343}\)

The overall time factor in the problem is the sum of the time it takes for the rock to reach the water and the time for the sound to travel back. It's crucial to understand how different speeds of motion, rock falling, and sound returning, combine to complete the observed event time.

Free Fall

Free fall refers to the motion of an object falling solely under the influence of gravity. In this context, the rock is in free fall as soon as it is dropped into the well. No other forces, like air resistance, are considered in basic free fall physics problems, simplifying our analysis and allowing us to focus strictly on the gravitational effect.
During free fall, objects accelerate downwards at the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)). As the rock falls, it gains speed until it hits the water. The time it takes for the rock to fall to the water's surface (\(t\)) can be calculated using the formula:

• \(t = \sqrt{\frac{2d}{g}}\)

This shows the direct relationship between fall time and distance, meaning the further it falls, the longer it takes, accelerating as it continues its descent.

Gravity

Gravity is the natural force that pulls objects towards the Earth's center, providing the acceleration necessary for free-fall motion. The constant value of \(9.8 \, \text{m/s}^2\) is pivotal in all problems dealing with motion under gravity close to the Earth’s surface.

• This problem relies on gravity's consistent acceleration to predict the motion of the rock.
• It's critical in understanding how objects behave when dropped, how they accelerate, and how their velocity increases over time.

While variables such as air resistance or initial velocity could complicate real-world scenarios, in such textbook exercises, gravity's uniform acceleration is assumed. This ensures calculations remain straightforward, focusing purely on the aspect of gravitational acceleration affecting the motion.