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Chapter 4: Problem 22

A small ball rolls horizontally off the edge of a tabletop that is \(1.20 \mathrm{~m}\) high. It strikes the floor at a point \(1.52 \mathrm{~m}\) horizontally from the table edge. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table?

Short Answer

Expert verified
(a) 0.495 seconds, (b) 3.07 m/s

Step by step solution

01

Analyze Vertical Motion

For vertical motion, the only force acting on the ball is gravity. We use the equation for free fall motion: \[ h = \frac{1}{2}gt^2 \]where \( h = 1.20 \text{ m} \) and \( g = 9.8 \text{ m/s}^2 \). Rearrange this equation to solve for \( t \) (the time in seconds) as follows:\[ t = \sqrt{\frac{2h}{g}} \]
02

Calculate Time in the Air

Substitute the given values into the rearranged formula:\[ t = \sqrt{\frac{2 \times 1.20 \text{ m}}{9.8 \text{ m/s}^2}} \approx \sqrt{0.2449} \approx 0.495 \text{ s} \]So, the ball is in the air for approximately 0.495 seconds.
03

Analyze Horizontal Motion

Since there is no horizontal acceleration (assuming air resistance is negligible), the horizontal motion can be analyzed using:\[ d = vt \]where \( d = 1.52 \text{ m} \) and \( t = 0.495 \text{ s} \). We need to solve for \( v \), which is the speed at which the ball leaves the table.
04

Calculate Initial Horizontal Speed

Rearrange the formula for horizontal motion to solve for \( v \) (the speed at which the ball leaves the table):\[ v = \frac{d}{t} = \frac{1.52 \text{ m}}{0.495 \text{ s}} \approx 3.07 \text{ m/s} \]Hence, the speed of the ball at the instant it leaves the table is approximately 3.07 m/s.

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Key Concepts

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

Horizontal and Vertical Motion
When dealing with projectile motion like that of a ball rolling off a table, it is essential to understand how both horizontal and vertical motions work. These two components occur simultaneously but can be analyzed independently.
  • Horizontal Motion: The ball moves sideways without any horizontal forces acting on it (assuming no air resistance). This means that its horizontal velocity remains constant throughout the fall.
  • Vertical Motion: The ball is affected by gravity, causing it to accelerate downwards. This part of the motion can be described as a free fall.
Understanding these components separately helps predict where a projectile will land. The time of flight and initial horizontal speed are crucial in solving such problems.
Free Fall
Free fall describes an object's motion under the influence of gravity alone. When the ball leaves the table, it instantly begins to free-fall because the only force acting on it is gravity.

Use the formula for vertical motion under constant acceleration due to gravity: \[ h = \frac{1}{2}gt^2 \]
  • \( h \): the height from which it falls (1.20 m in this case)
  • \( g \): the acceleration due to gravity (approximately 9.8 m/s虏)
  • While the horizontal motion may continue at a constant speed, the vertical motion will increase in speed as it gets closer to the ground.
The time it takes to fall can be determined by rearranging the equation and solving for \( t \). Understanding this helps predict how long an object will be in the air before hitting the ground.
Kinematics
Kinematics is the study of motion without considering the forces that cause these motions. It allows us to describe the motion of objects using equations.

In the projectile motion of the ball, kinematic equations help determine unknown variables like time in the air and initial speed.
  • Horizontal and vertical components use different kinematic principles. The equations consider distances, speeds, and times to solve for unknowns.
  • In this scenario, the horizontal velocity remains constant, while the vertical component uses acceleration due to gravity to determine the time in the air.
  • The independence of horizontal and vertical motions is a major concept in kinematics, allowing us to solve problems in steps as demonstrated in the solution provided.
Initial Velocity
Initial velocity is a critical parameter in projectile motion problems. It refers to the speed at which an object begins its motion.

In the problem of the ball rolling off the table, the initial velocity is entirely in the horizontal direction. The ball does not initially move vertically as it leaves the table.
  • The initial vertical velocity is zero, while the horizontal velocity can be calculated using the formula: \[ v = \frac{d}{t} \]
  • This equation helps determine how fast the ball was moving horizontally when it left the table to reach the floor at a distance of 1.52 m in 0.495 seconds.
  • By understanding initial velocity, you can predict the full trajectory and final position of any projectile object based on the speed and direction it starts with.
Knowing the initial velocities helps in making accurate calculations regarding the future motion of the object.

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Most popular questions from this chapter

An electron's position is given by \(\vec{r}=3.00 t \hat{\mathrm{i}}-\) \(4.00 t^{2} \hat{\mathrm{j}}+2.00 \hat{\mathrm{k}}\), with \(t\) in seconds and \(\vec{r}\) in meters. (a) In unit-vector notation, what is the electron's velocity \(\vec{v}(t)\). At \(t=2.00 \mathrm{~s}\), what is \(\vec{v}\) (b) in unit- vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the \(x\) axis?

A projectile is fired horizontally from a gun that is \(45.0 \mathrm{~m}\) above flat ground, emerging from the gun with a speed of \(250 \mathrm{~m} / \mathrm{s}\). (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

An electron having an initial horizontal velocity of magnitude \(1.00 \times 10^{9} \mathrm{~cm} / \mathrm{s}\) travels into the region between two horizontal metal plates that are electrically charged. In that region, the electron travels a horizontal distance of \(2.00 \mathrm{~cm}\) and has a constant downward acceleration of magnitude \(1.00 \times 10^{17} \mathrm{~cm} / \mathrm{s}^{2}\) due to the charged plates. Find (a) the time the electron takes to travel the \(2.00 \mathrm{~cm},(\mathrm{~b})\) the vertical distance it travels during that time, and the magnitudes of its (c) horizontal and (d) vertical velocity components as it emerges from the region.

At one instant a bicyclist is \(40.0 \mathrm{~m}\) due east of a park's flagpole, going due south with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). Then \(30.0\) s later, the cyclist is \(40.0 \mathrm{~m}\) due north of the flagpole, going due east with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). For the cyclist in this \(30.0\) s interval, what are the (a) magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity, and the (c) magnitude and (f) direction of the average acceleration?

A wooden boxcar is moving along a straight railroad track at speed \(v_{1}\), A sniper fires a bullet (initial speed \(v_{2}\) ) at it from a highpowered rifle. The bullet passes through both lengthwise walls of the car, its entrance and exit holes being exactly opposite each other as viewed from within the car. From what direction, relatiye to the track, is the bullet fired? Assume that the bullet is not deflected upon entering the car, but that its speed decreases by \(20 \%\). Take \(v_{1}=85 \mathrm{~km} / \mathrm{h}\) and \(v_{2}=650 \mathrm{~m} / \mathrm{s}\). (Why don't you need to know the width of the boxcar?)
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