/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 A ball is dropped from rest from... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A ball is dropped from rest from the top of a cliff that is \(24 \mathrm{~m}\) high. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is exactly the same as that with which the first ball eventually hits the ground. In the absence of air resistance, the motions of the balls are just the reverse of each other. Determine how far below the top of the cliff the balls cross paths.

Short Answer

Expert verified
They cross paths about 18.0 m below the top of the cliff.

Step by step solution

01

Find the final speed of the first ball

The first ball is dropped from a height of 24 meters with an initial velocity of 0 m/s. Using the equation for the final velocity in free fall, you can find the speed at which the ball hits the ground:\[ v^2 = u^2 + 2gh \]where \( u = 0 \), \( g = 9.8 \ m/s^2 \), and \( h = 24 \ m \).\[ v = \sqrt{2 \times 9.8 \times 24} \approx 21.7 \ m/s \]
02

Determine equations of motion for each ball

For the first ball dropped from the cliff, its position as a function of time can be described as:\[ y_1 = 24 - \frac{1}{2}gt^2 \]For the second ball thrown upwards, with initial velocity \( v = 21.7 \ m/s \), its position as a function of time is:\[ y_2 = v \cdot t - \frac{1}{2}gt^2 = 21.7 \cdot t - \frac{1}{2} \cdot 9.8 \cdot t^2 \]
03

Set the equations of motion equal to find crossing point

Set the position equations equal to each other to find the time \( t \) at which both balls are at the same height:\[ 24 - \frac{1}{2}gt^2 = 21.7t - \frac{1}{2}gt^2 \]This simplifies to \( 24 = 21.7t \).Solving for \( t \), we find:\[ t = \frac{24}{21.7} \approx 1.107 \ seconds \]
04

Calculate the position of the balls at time t

Substitute \( t \) back into one of the position equations to find how far below the top of the cliff the balls cross paths. We'll use the equation for the first ball's position:\[ y_1 = 24 - \frac{1}{2} \times 9.8 \times (1.107)^2 \]\[ y_1 = 24 - \frac{1}{2} \times 9.8 \times 1.225 \approx 24 - 5.993 \approx 18.007 \ m \]Thus, the balls cross about \( 18.007 \ m \) below the top of the cliff.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Free Fall
Free fall is an essential concept in physics, describing the motion of objects as they are influenced solely by gravity. When an object is in free fall, it experiences an acceleration due to Earth's gravity, approximately equal to 9.8 m/s². This acceleration causes the object to increase in speed as it descends, regardless of its mass. For the first ball in our exercise, which is dropped from the cliff, this motion begins with an initial velocity of 0 m/s since it is simply released, not thrown.

In free fall, some key attributes include:
  • The only force acting on the object is gravity.
  • Air resistance is typically ignored in simplifying calculations.
  • Motion can be consistently described using specific kinematic equations.
This pure gravitational motion allows us to calculate how long it will take an object to hit the ground and at what speed, making free-fall problems both interesting and rich with real-world applications.
Kinematics Equations
Kinematics equations are mathematical formulas used to predict the future motion of objects. They apply to systems themselves accelerating at a constant rate. In our problem, these equations are essential to understand the motion paths of both balls.
  • The equation to find the final velocity of an object in free fall is given by: \[ v^2 = u^2 + 2gh \]
  • To describe the position of an object over time, we use:\[ y_1 = 24 - \frac{1}{2}gt^2 \]for the first ball, and \[ y_2 = v \cdot t - \frac{1}{2}gt^2 \]for the second ball.
These kinematic equations function as a toolkit for solving everyday motion problems. They are particularly useful because they allow us to avoid complex calculations involving force and mass by focusing on displacement, velocity, and acceleration. This makes kinematic equations a favorite tool for students tackling projectile motion or free-fall scenarios.
Velocity
Velocity is a vector quantity that describes both the speed and direction of an object's motion. In this exercise, understanding velocity is crucial for determining when and where the two balls will meet on their paths.

For the first ball dropped from the cliff, its velocity increase is due solely to gravitational acceleration. After falling 24 meters, its velocity is approximately 21.7 m/s just before impact. Meanwhile, the second ball starts its journey with an upward velocity equal to this final value, creating symmetry in the scenario.

Some important features of velocity to remember are:
  • It's always directional, which means it can be positive (e.g., upward) or negative (e.g., downward).
  • It changes over time if an object accelerates or decelerates, as seen in this problem where velocity shifts from initial values determined by gravity.
  • When solving physics problems, calculating velocity at different time points helps predict future positions, like where the paths of the two balls intersect.
By understanding velocity in context, it becomes easier to solve not only textbook problems but also real-life physics challenges.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Interactive Solution \(\underline{2.31}\) at offers help in modeling this problem. A car is traveling at a constant speed of \(33 \mathrm{~m} / \mathrm{s}\) on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is \(2.5 \mathrm{~km}\) away?

From her bedroom window a girl drops a water-filled balloon to the ground, \(6.0 \mathrm{~m}\) below. If the balloon is released from rest, how long is it in the air?

A wrecking ball is hanging at rest from a crane when suddenly the cable breaks. The time it takes for the ball to fall halfway to the ground is \(1.2 \mathrm{~s}\). Find the time it takes for the ball to fall from rest all the way to the ground.

A jogger accelerates from rest to \(3.0 \mathrm{~m} / \mathrm{s}\) in \(2.0 \mathrm{~s}\). A car accelerates from 38.0 to \(41.0 \mathrm{~m} / \mathrm{s}\) also in \(2.0 \mathrm{~s}\). (a) Find the acceleration (magnitude only) of the jogger. (b) Determine the acceleration (magnitude only) of the car. (c) Does the car travel farther than the jogger during the \(2.0 \mathrm{~s}\) ? If so, how much farther?

The Space Shuttle travels at a speed of about \(7.6 \times 10^{3} \mathrm{~m} / \mathrm{s}\). The blink of an astronaut's eye lasts about \(110 \mathrm{~ms}\). How many football fields (length \(=91.4 \mathrm{~m}\) ) does the Shuttle cover in the blink of an eye?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.