/*! 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 15 After falling from rest from a h... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 15

After falling from rest from a height of \(30 \mathrm{~m}\), a \(0.50-\mathrm{kg}\) ball rebounds upward, reaching a height of \(20 \mathrm{~m}\). If the contact between ball and ground lasted \(2.0 \mathrm{~ms}\), what average force was exerted on the ball?

Short Answer

Expert verified
-11062.5 N

Step by step solution

01

Identify initial velocity after falling

The initial velocity of the ball right after falling can be obtained using the equation of motion \(v^2 = u^2 + 2as\), where \(u\) is the initial velocity (0), \(a\) is the acceleration due to gravity (9.8 m/s虏), and \(s\) is the distance fallen (30 m). Thus the velocity after falling is \(\sqrt{2 * 9.8 * 30} \approx 24.25 \, m/s\).
02

Identify final velocity after rebound

The final velocity before the ball starts to rise again can be obtained with the same equation of motion as before where \(s\) is now the height it reached after rebounding (20 m). We have \(v^2 = u^2 + 2as\), hence the velocity after rebound is \(-\sqrt{2 * 9.8 * 20} \approx -20 \, m/s\). The negative sign indicates that the direction of the velocity is upward.
03

Calculate change in velocity

The change in velocity is identified by subtracting the final velocity from the initial velocity. This gives \(-20 \, m/s - 24.25 \, m/s = -44.25 \, m/s\). The negative sign indicates that the velocity is in the opposite direction of the initial velocity.
04

Calculate average force

Finally, we use the second law of motion, \(F = ma\), where \(m\) is the mass (0.50 kg) and \(a\) is the acceleration. Here, acceleration is essentially the change in velocity over the change in time. The time of contact given is 2.0 ms. So, the average force exerted on the ball is \(F = m * \frac{\Delta v}{\Delta t} = 0.50 * \frac{-44.25}{2.0 * 10^{-3}} = -11062.5 \, N\). The negative sign indicates that the force is directed upward.
05

Conclusion

The average force exerted on the ball is 11062.5 N. This force is in the opposite direction to the force of gravity, hence it is exerted upward.

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 91影视!

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

A boy coasts down a hill on a sled, reaching a level surface at the bottom with a speed of \(7.0 \mathrm{~m} / \mathrm{s}\). If the coefficient of friction between the sled's runners and the snow is \(0.050\) and the boy and sled together weigh \(600 \mathrm{~N}\), how far does the sled travel on the level surface before coming to rest?

As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). In the process, he moves his hand through a distance of \(1.50 \mathrm{~m}\). If the ball has a mass of \(0.150 \mathrm{~kg}\), find the force he exerts on the ball to give it this upward speed.

The distance between two telephone poles is \(50.0 \mathrm{~m}\). When a \(1.00-\mathrm{kg}\) bird lands on the telephone wire midway between the poles, the wire sags \(0.200 \mathrm{~m}\). Draw a free-body diagram of the bird. How much tension does the bird produce in the wire? Ignore the weight of the wire.

A crate of mass \(m=\) \(32 \mathrm{~kg}\) rides on the bed of a truck attached by a cord to the back of the cab as in Figure \(\mathrm{P} 4.26 .\) The cord can withstand a maximum tension of \(68 \mathrm{~N}\) before breaking. Neglecting friction between the crate and truck bed, find the maximum acceleration the truck can have before the cord breaks.

A \(5.0-g\) bullet leaves the muzzle of a rifle with a speed of \(320 \mathrm{~m} / \mathrm{s}\). What force (assumed constant) is exerted on the bullet while it is traveling down the \(0.82\)-m-long barrel of the rifle?
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Magnetism

Read Explanation

Famous Physicists

Read Explanation

Particle Model of Matter

Read Explanation

Kinematics Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation
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.