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At a construction site a pipe wrench struck the ground with a speed of \(24 \mathrm{~m} / \mathrm{s}\). (a) From what height was it inadvertently dropped? (b) How long was it falling? (c) Sketch graphs of \(y, v\), and \(a\) versus \(t\) for the wrench.

Short Answer

Expert verified
The pipe wrench fell from a height of 29.4 m and was falling for 2.45 s.

Step by step solution

01

Identify the Known Values

We know the final velocity of the wrench when it hits the ground was \( v = 24 \text{ m/s} \), and the acceleration due to gravity is \( g = 9.8 \text{ m/s}^2 \). We need to find the initial height \( h \), assuming initial velocity \( u = 0 \).
02

Use Kinematic Equation to Find Height

Using the kinematic equation \( v^2 = u^2 + 2gh \), where \( u=0 \), we have: \[ v^2 = 2gh \]. Substitute \( v = 24 \text{ m/s} \) and \( g = 9.8 \text{ m/s}^2 \) to find \( h \): \[ (24)^2 = 2 \times 9.8 \times h \]. This simplifies to \( h = \frac{576}{19.6} \approx 29.4 \text{ m} \).
03

Use Kinematic Equation to Find Time of Fall

Using the kinematic equation \( v = u + gt \), where \( u=0 \), we have: \[ v = gt \]. Substitute \( v = 24 \text{ m/s} \) and \( g = 9.8 \text{ m/s}^2 \) to solve for \( t \): \[ 24 = 9.8t \]. This gives \( t = \frac{24}{9.8} \approx 2.45 \text{ seconds} \).
04

Sketch Graphs

For graph (a) \( y(t) \): The position as a function of time, \( y(t) = h - \frac{1}{2}gt^2 \). It's a downward opening parabola starting from \( h \). For graph (b) \( v(t) \): The velocity vs. time is a linear function, \( v(t) = gt \), starting from zero and increasing. For graph (c) \( a(t) \): The acceleration vs. time graph is a horizontal line \( a(t) = g \), as acceleration due to gravity is constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematic Equations
Kinematic equations are crucial tools for solving problems related to motion in physics. They describe how position, velocity, and acceleration are related to time. These equations help us predict the position and velocity of an object when specific variables, like acceleration or time, are known. There are four main kinematic equations to consider:
  • Equation 1: \( v = u + at \) - relates final velocity \( v \), initial velocity \( u \), acceleration \( a \), and time \( t \).
  • Equation 2: \( s = ut + \frac{1}{2}at^2 \) - calculates displacement \( s \), considering initial velocity, time, and acceleration.
  • Equation 3: \( v^2 = u^2 + 2as \) - useful for finding the final velocity squared, initial velocity squared, acceleration, and displacement.
  • Equation 4: \( s = vt - \frac{1}{2}at^2 \) - another form to calculate displacement when initial velocity is zero.
These equations assume motion is in a straight line with constant acceleration, making them perfect for free-fall problems where only gravity acts on the object.
Free Fall
Free fall refers to the motion of objects falling solely under the influence of gravity, without any resistance from air or other forces. In such cases, gravitational acceleration is the only acceleration affecting the object. Free fall is an excellent example of motion with constant acceleration, making it a prime candidate for using kinematic equations.
  • An important feature of free fall is that it is independent of the object's mass or shape. In a vacuum, a feather and a hammer will hit the ground simultaneously if dropped from the same height, as famously demonstrated by astronaut David Scott on the Moon.
  • For objects in free fall near Earth's surface, the acceleration due to gravity is approximately \( 9.8 \text{ m/s}^2 \).
In our example, the pipe wrench is experiencing free fall because its falling motion is solely influenced by gravity until it hits the ground.
Acceleration Due to Gravity
The acceleration due to gravity, often denoted as \( g \), is the rate at which an object accelerates when falling freely towards Earth's surface. This acceleration is a constant value near Earth's surface, approximately \( 9.8 \text{ m/s}^2 \). The concept of constant acceleration is critical because it simplifies calculations involving motion, particularly in free-fall scenarios.
  • For any object in free fall, this value is applied in kinematic equations to calculate other aspects of motion, like velocity and time elapsed.
  • However, the exact value of \( g \) can slightly vary based on location on Earth's surface due to factors like altitude and Earth's non-uniform shape.
Using \( g \), we can determine that the pipe wrench accelerating at \( 9.8 \text{ m/s}^2 \) will follow predictable motion paths, making calculations about its fall straightforward and manageable.
Velocity-Time Graph
A velocity-time graph depicts how the velocity of an object changes over time. For our problem, considering the free fall of the pipe wrench, the graph shows a linear relationship because the acceleration (gravity in this case) is constant.
  • The slope of the line is equal to the acceleration due to gravity, \( 9.8 \text{ m/s}^2 \). This means the velocity of the pipe wrench increases linearly as time passes.
  • In the provided problem, the initial velocity is zero, so the graph starts at the origin. As time increases, velocity increases linearly, representing a straight line sloping upwards.
This kind of graph is helpful to visualize and further predict how long it will take an object in free fall to reach a specific velocity, like how it was used to determine the pipe wrench's velocity when it hit the ground.
Position-Time Graph
A position-time graph shows how the position or height of an object changes over time. For an object in free fall, like the pipe wrench, the graph is a downward-opening parabola because the initial position is reduced at a rate that increases with time.
  • The curve starts from the maximum height (where the object was initially dropped) and descends as time progresses, depicting the accelerating descent under gravity.
  • The downward curve indicates increasingly faster movement towards the ground, which is characteristic of the increasing velocity in free fall.
In this situation, the graph helps illustrate the journey from the drop height to the ground, emphasizing how position changes over the fall duration due to gravitational acceleration.

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Most popular questions from this chapter

A shuffleboard disk is accelerated at a constant rate from rest to a speed of \(6.0 \mathrm{~m} / \mathrm{s}\) over a \(1.8 \mathrm{~m}\) distance by a player using a cue. At this point the disk loses contact with the cue and slows at a constant rate of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) until it stops. (a) How much time elapses from when the disk begins to accelerate until it stops? (b) What total distance does the disk travel?

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