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Chapter 2: Problem 49

A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 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) 30.66 m, (b) 24.53 m/s

Step by step solution

01

Understand the problem

We have a brick dropped from the roof of a building with no initial velocity. The brick falls for 2.50 seconds without air resistance. We need to find the height of the building, the final velocity of the brick, and sketch graphs for acceleration, velocity, and position with respect to time.
02

Identify known values

From the problem, the following values are known:- Initial velocity (\( v_0 \)) = 0 m/s- Time of fall (\( t \)) = 2.50 s- Acceleration due to gravity (\( g \)) = 9.81 m/s虏.
03

Calculate the height of the building

Use the equation for distance fallen under constant acceleration: \[ h = v_0t + \frac{1}{2}gt^2 \]Substitute the known values:\[ h = 0 \times 2.50 + \frac{1}{2} \times 9.81 \times (2.50)^2 \]Calculate:\[ h = \frac{1}{2} \times 9.81 \times 6.25 = 30.65625 \text{ meters} \]
04

Determine the final velocity before impact

Use the equation for final velocity:\[ v = v_0 + gt \]Substitute the known values:\[ v = 0 + 9.81 \times 2.50 \]Calculate:\[ v = 24.525 \text{ m/s} \]
05

Sketch the graphs

1. **Acceleration vs. Time**: The graph is a horizontal line at 9.81 m/s虏 because acceleration due to gravity is constant. 2. **Velocity vs. Time**: The graph is a straight line with a positive slope passing through (0,0) and ending at (2.5, 24.525 m/s), representing constant acceleration. 3. **Position vs. Time**: The graph is a parabola whose vertex is at the origin, going downward, reflecting the increasing displacement over time.

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Key Concepts

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

Acceleration Due to Gravity
When objects are in free fall, they are pulled towards Earth by a constant force called gravity. This force causes the acceleration of the object, which is standardized as approximately 9.81 meters per second squared (m/s虏) on Earth's surface.
This means that every second, an object's velocity will increase by about 9.81 m/s when it is in free fall.
The acceleration due to gravity is symbolized by the letter "g" in physics equations. In our problem, this constant acceleration is crucial for calculating the brick's final speed and how far it falls.
  • Constant: 9.81 m/s虏
  • Direction: Towards Earth's center
  • Induces velocity change
This value is especially important in free fall problems since it remains the same regardless of the object's mass or shape.
Kinematic Equations
Kinematic equations are essential in solving motion problems and relate displacement, velocity, acceleration, and time.
For an object in uniform acceleration, like our falling brick, kinematic equations allow us to calculate unknown variables.
One primary kinematic equation used here is:
  • \[ h = v_0t + \frac{1}{2}gt^2 \]
This equation calculates the height (or displacement) "h" an object falls in time "t" with an initial velocity "v_0" and constant acceleration "g". Another important equation is:
  • \[ v = v_0 + gt \]
Used to find final velocity "v" after time "t".
The beauty of these equations is their ability to interconnect speed, time, distance, and acceleration.
Final Velocity
The final velocity of an object in free fall is its speed just before it hits the ground.
To determine this, we use the kinematic equation for velocity:
  • \[ v = v_0 + gt \]
Given that the initial velocity "v_0" is zero (because the brick is dropped and not thrown), the final velocity simplifies to:\[ v = gt \]
Substituting our values:
\[ v = 9.81 \times 2.50 \]
This results in a final velocity of 24.525 m/s.
This implies that the brick's speed increases steadily due to gravity over the 2.5 seconds it falls.
Displacement
Displacement refers to how far and in what direction an object has moved from its starting position.
In free fall, displacement can be calculated with the kinematic equation:
  • \[ h = \frac{1}{2}gt^2 \]
In the context of our falling brick, displacement helps us find the building's height from which the brick was dropped. Plugging in the values gives:
\[ h = \frac{1}{2} \times 9.81 \times (2.50)^2 \]
Calculating this yields approximately 30.65625 meters.
This displacement represents how much the brick moves vertically downward from its initial position on the roof. It reflects the increasing speed over time due to gravitational pull.

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Most popular questions from this chapter

Life from space? One rather controversial suggestion for the origin of life on the earth is that the seeds of life arrived here from outer space in tiny spores-a theory called panspermia. Of course, any life in such spores would have to survive a long journey through the frigid near vacuum of outer space. If these spores originated at the distance of our nearest star, Alpha Centauri, and traveled at a constant \(1000 \mathrm{km} / \mathrm{s},\) how many years would it take for them to make the journey here? (Alpha Centauri is approximately 4\(\frac{1}{4}\) light years away, and the light year is the distance light travels in 1 year.)

Earthquake waves. Earthquakes produce several types of shock waves. The best known are the P-waves (P for primary or pressure) and the S-waves (S for secondary or shear). In the earth's crust, P-waves travel at around 6.5 \(\mathrm{km} / \mathrm{s}\) while S-waves move at about 3.5 \(\mathrm{km} / \mathrm{s}\) . (The actual speeds vary with the type of material the waves are going through.) The time delay between the arrival of these two types of waves at a seismic recording station tells geologists how far away the earthquake that produced the waves occurred. (a) If the time delay at a seismic station is 33 s, how far from that station did the earthquake that produced the waves occurred. (a) If the time delay at a seismic station is 33 s, how far from that station did the earthquake occur? (b) One form of earthquake warning system detects the faster (but less damaging) P-waves and sounds an alarm when they first arrive, giving people a short time to seek cover before the more dangerous S-waves arrive. If an earthquake occurs 375 \(\mathrm{km}\) away from such a warning device, how much time would people have to take cover between the alarm and the arrival of the S-waves?

A rock is thrown vertically upward with a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) from the roof of a building that is 60.0 \(\mathrm{m}\) above the ground. (a) In how many seconds after being thrown does the rock strike the ground? (b) What is the speed of the rock just before it strikes the ground? Assume free fall.

Two rockets having the same acceleration start from rest, but rocket \(A\) travels for twice as much time as rocket \(B\) . If rocket \(A\) goes a distance of \(250 \mathrm{km},\) how far will rocket \(B\) go? (b) If rocket \(A\) reaches a speed of \(350 \mathrm{m} / \mathrm{s},\) what speed will rocket \(B\) reach?

Freeway ramps. In designing a freeway on-ramp, you must make it long enough for cars to be able to accelerate and merge safely with traffic traveling at speeds of 70 mph. For an off-ramp, the cars must be able to come to a stop from 70 mph over the length of the ramp. Your traffic engineers provide the following data: Powerful cars can accelerate from rest to 60 mph in 5.0 s, while less powerful cars take twice as long to do this. When slowing down from 60 mph, a car with good tire treads will take 12.0 s to stop, whereas one with bald tires takes 20.0 s. In order to safely accommodate all the types of vehicles, what should be the minimum length (in meters) of on-ramps and off- ramps? Assume constant acceleration.
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