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Chapter 2: Problem 59

A student drops a ball from the top of a tall building; the ball takes 2.8 s to reach the ground. (a) What was the ball's speed just before hitting the ground? (b) What is the height of the building?

Short Answer

Expert verified
The ball's speed before hitting the ground is 27.44 m/s; the height of the building is 38.416 m.

Step by step solution

01

Define the known variables

We'll begin by noting the known values from the problem. The initial velocity \( v_i \) is 0 m/s (since the ball is dropped, not thrown), the acceleration \( a \) is the acceleration due to gravity \( g \), which is approximately 9.8 m/s虏, and the time \( t \) is 2.8 seconds.
02

Calculate the final velocity

To find the final velocity \( v_f \) of the ball just before it hits the ground, we use the equation of motion: \( v_f = v_i + a \, t \).Substituting the known values, \( v_f = 0 + 9.8 \, \text{m/s}^2 \, \times \, 2.8 \, \text{s} = 27.44 \, \text{m/s} \).
03

Calculate the height of the building

The height \( h \) of the building can be found using the equation \( h = v_i \, t + \frac{1}{2} a \, t^2 \).Substitute the known values: \( h = 0 + \frac{1}{2} \times 9.8 \, \text{m/s}^2 \, \times (2.8 \, \text{s})^2 = 38.416 \, \text{m} \).

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Key Concepts

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

Free Fall
Imagine dropping a ball from your hand off the edge of a tall building. This kind of motion is called free fall. In free fall, the only force acting on the object is gravity. There is no initial push or throw, just a simple drop. This absence of air resistance, especially in theory or close approximations, allows the object to accelerate towards the Earth unimpeded.
When something is in free fall:
  • The initial velocity is zero since no initial force is applied vertically.
  • The object will continuously speed up as it falls due to gravity.
  • The object's motion can be predicted using specific motion equations.
Understanding these characteristics is crucial for solving problems related to falling objects.
Acceleration Due to Gravity
Gravity is an unstoppable force that pulls all objects toward the center of the Earth, accelerating them at a constant rate when in free fall. This is known as the "acceleration due to gravity," often represented as "g." On Earth, this value is approximately 9.8 m/s虏, although it can vary slightly depending on geographical location.

Here's what you need to know about it:
  • This acceleration is independent of the object's mass. All objects in free fall near Earth's surface will accelerate at the same rate, ignoring air resistance.
  • It acts constantly in the downward direction, meaning it will continually increase the velocity of a falling object over time.
  • This constant acceleration forms the base of many equations used to calculate motion during free fall.
By understanding this concept, we can accurately determine various aspects of an object's motion as it falls.
Equations of Motion
The equations of motion are powerful tools in physics that help predict and describe the movement of objects. In the context of free fall, these equations become particularly useful. There are a few key equations that apply when considering objects in free fall:

To find the final speed just before the object hits the ground, the equation:
  • \( v_f = v_i + a \times t \)
  • With \( v_i = 0 \), \( a = 9.8 \, \text{m/s}^2 \), and \( t = 2.8 \, \text{s} \), we find the speed to be \( 27.44 \, \text{m/s} \).
To find the distance or height traveled by the object, use:
  • \( h = v_i \times t + \frac{1}{2} \times a \times t^2 \)
  • Given the same known values, the height comes out to \( 38.416 \, \text{m} \).
Using these equations, we can not only solve textbook problems but also gain an intuitive understanding of how objects behave when free-falling under gravity's influence.

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Most popular questions from this chapter

A tennis ball is dropped from a height of \(10.0 \mathrm{~m}\). It rebounds off the floor and comes up to a height of only \(4.00 \mathrm{~m}\) on its first rebound. (Ignore the small amount of time the ball is in contact with the floor.) (a) Determine the ball's speed just before it hits the floor on the way down. (b) Determine the ball's speed as it leaves the floor on its way up to its first rebound height. (c) How long is the ball in the air from the time it is dropped until the time it reaches its maximum height on the first rebound?

A train normally travels at a uniform speed of \(72 \mathrm{~km} / \mathrm{h}\) on a long stretch of straight, level track. On a particular day, the train must make a 2.0 -min stop at a station along this track. If the train decelerates at a uniform rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) and, after the stop, accelerates at a rate of \(0.50 \mathrm{~m} / \mathrm{s}^{2},\) how much time is lost because of stopping at the station?

A car traveling at a speed of \(v\) can brake to an emergency stop in a distance \(x\). Assuming all other driving conditions are similar, if the traveling speed of the car doubles, the stopping distance will be (1) \(\sqrt{2 x,}\) (2) \(2 x\) (3) \(4 x\). (b) A driver traveling at \(40.0 \mathrm{~km} / \mathrm{h}\) in a school zone can brake to an emergency stop in \(3.00 \mathrm{~m}\). What would be the braking distance if the car were traveling at \(60.0 \mathrm{~km} / \mathrm{h} ?\)

For the motion of a dropped object in free fall, sketch the general forms of the graphs of (a) \(v\) versus \(t\) and (b) \(y\) versus \(t\)

A car accelerates from rest at a constant rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) for \(5.0 \mathrm{~s}\). (a) What is the speed of the car at the end of that time? (b) How far does the car travel in this time?
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