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A brick is released with no initial speed from the roof of a building and strikes the ground in \(2.50 \mathrm{~s},\) encountering no appreciable air drag. (a) How tall, in meters, is the building? (b) How fast is the brick moving just before it reaches the ground? (c) Sketch graphs of this falling brick's acceleration, velocity, and vertical position as functions of time.

Short Answer

Expert verified
(a) 30.64 m; (b) 24.53 m/s; (c) See graphs for acceleration, velocity, and position.

Step by step solution

01

Understand the Initial Conditions

The brick is released from a building with no initial speed, which means its initial velocity is 0 m/s. It is experiencing free fall under gravity without air resistance. The time taken to hit the ground is given as 2.50 s.
02

Calculate the Height of the Building

Use the formula \( h = \frac{1}{2}gt^2 \), where \( g \) is the acceleration due to gravity (9.81 m/s²) and \( t \) is the time (2.50 s). Substitute these values:\[ h = \frac{1}{2} \times 9.81 \times (2.50)^2 = 30.6375 \text{ meters} \]Thus, the height of the building is approximately 30.64 meters.
03

Calculate the Final Velocity Before Impact

Use the formula \( v = gt \), where \( g \) is 9.81 m/s² and \( t \) is 2.50 s. Calculate as follows:\[ v = 9.81 \times 2.50 = 24.525 \text{ m/s} \]So, the final velocity of the brick just before it hits the ground is approximately 24.53 m/s.
04

Graphical Representation

(a) **Acceleration Graph**: Acceleration is constant at -9.81 m/s² (since downward is negative in this context). The graph is a horizontal line. (b) **Velocity Graph**: Starts at 0 m/s and becomes increasingly negative (downward) over time, forming a straight line with slope -9.81. (c) **Position Graph**: Starts at 0 and curves downward, forming a parabolic shape as it represents displacement over time due to constant acceleration.

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Key Concepts

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

Acceleration Due to Gravity
In physics, when an object falls freely under the influence of gravity alone, it experiences a specific and constant acceleration. This is referred to as the "acceleration due to gravity" and commonly denoted by the symbol \( g \). On Earth, the standard value of \( g \) is approximately 9.81 m/s².

This means that an object in free fall will increase its velocity by 9.81 meters per second every second, assuming there is no air resistance. The direction of this acceleration is towards the center of the Earth, which is typically considered negative if you choose down as the negative direction in your coordinate system.

This constant acceleration is the same for all objects, regardless of their mass, and is a crucial factor in solving free fall problems.
Velocity Calculation
In free fall, calculating the velocity of an object just before it impacts the ground involves knowing the time it has been falling and the acceleration due to gravity. The key formula we use is:

\[ v = gt \]

Here, \( v \) is the final velocity, \( g \) is the acceleration due to gravity (9.81 m/s²), and \( t \) is the time the object has been falling. In our exercise, with a time of 2.50 seconds, the final velocity just before hitting ground is calculated as 24.53 m/s.

This velocity implies how fast the object was traveling at the moment of impact, and solving such problems helps in understanding motion under constant acceleration.
Kinematics
Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. During free fall, kinematics allows us to describe and predict the position and velocity of an object.

Crucial equations of kinematics for uniformly accelerated motion include:
  • \( v = u + gt \)
  • \( s = ut + \frac{1}{2}gt^2 \)
  • \( v^2 = u^2 + 2gs \)

In these equations, \( u \) represents initial velocity (which is zero in our problem), \( v \) is final velocity, \( s \) is the displacement, and \( t \) is the time. By applying these equations, one can find parameters like the distance fallen and final velocity, as demonstrated with the brick.
Graphical Representation of Motion
Visualizing motion through graphs provides considerable insight into the changes in various parameters such as acceleration, velocity, and position over time. Let's discuss these in the context of the falling brick:

  • **Acceleration Graph:** This graph is straightforward, reflecting the constant acceleration due to gravity. It is a horizontal line at -9.81 m/s², indicating a steady rate of velocity change per second, directed downwards.
  • **Velocity Graph:** Starting at zero, this graph shows a straight line declining downward at a constant rate, capturing the linear increase in speed as it falls under gravity. This line's slope equals \( g \), showing constant acceleration.
  • **Position Graph:** As time progresses, the graph shows an exponentially decreasing curve. This parabolic shape reflects how the object's displacement increases at a changing rate due to the consistent gravitational pull, highlighting the quadratic relationship between time and distance fallen.

Graphing these attributes helps students understand the dynamic nature of motion influenced by gravity in a more tangible and visual manner.

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Most popular questions from this chapter

At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constant acceleration of \(2.50 \mathrm{~m} / \mathrm{s}^{2}\). At the same instant, a truck, traveling with a constant speed of \(15.0 \mathrm{~m} / \mathrm{s}\), overtakes and passes the automobile. (a) How far beyond its starting point does the automobile overtake the truck? (b) How fast is the automobile traveling when it overtakes the truck?

The Beretta Model \(92 \mathrm{~S}\) (the standard-issue U.S. army pistol) has a barrel \(127 \mathrm{~mm}\) long. The bullets leave this barrel with a muzzle velocity of \(335 \mathrm{~m} / \mathrm{s}\). (a) What is the acceleration of the bullet while it is in the barrel, assuming it to be constant? Express your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g^{\prime}\) s. (b) For how long is the bullet in the barrel?

One way to measure \(g\) on another planet or moon by remote sensing is to measure how long it takes an object to fall a given distance. A lander vehicle on a distant planet records the fact that it takes \(3.17 \mathrm{~s}\) for a ball to fall freely \(11.26 \mathrm{~m},\) starting from rest. (a) What is the acceleration due to gravity on that planet? Express your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and in earth \(g^{\prime} \mathrm{s}\). (b) How fast is the ball moving just as it lands?

While riding on a bus traveling down the highway, you notice that it takes 2 min to travel from one roadside mile marker to the next one. (a) What is your speed in \(\mathrm{mi} / \mathrm{h} ?\) (b) How long does it take the bus to travel 100 yds?

A car in the northbound lane is sitting at a red light. At the moment the light turns green, the car accelerates from rest at \(2 \mathrm{~m} / \mathrm{s}^{2}\). At this moment, there is also a car in the southbound lane that is \(200 \mathrm{~m}\) away and traveling at a constant \(25 \mathrm{~m} / \mathrm{s}\). The northbound car maintains its acceleration until the two cars pass each other. (a) How long after the light turns green do the cars pass each other? (b) How far from the red light are they when they pass each other?

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