/*! 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 1 Two athletes jump straight up. U... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 1

Two athletes jump straight up. Upon leaving the ground, Adam has half the initial speed of Bob. Compared to Adam, Bob jumps a) 0.50 times as high. b) 1.41 times as high. c) twice as high. d) three times as high. e) four times as high.

Short Answer

Expert verified
Answer: e) Bob jumps four times as high as Adam.

Step by step solution

01

1. Write down given information

We have the initial speed of Adam and Bob, with Adam having half the initial speed of Bob. Let Va be the initial speed of Adam and Vb be the initial speed of Bob. Then: Va = 0.5 * Vb
02

2. Use kinematic equations

We know that the height (h) depends on the initial speed (v0) of a jump. The maximum height is achieved when the final velocity is zero (vf=0). Using the kinematic equation: vf^2 = v0^2 - 2gh, we can find the height by rearranging it for h: h = (v0^2) / (2g) Where g is the acceleration due to gravity, which is approximately 9.81 m/s^2.
03

3. Substitute initial speeds

Now we can find the height jumped by both Adam and Bob using their initial speeds. For Adam: ha = (Va^2) / (2g) For Bob: hb = (Vb^2) / (2g)
04

4. Compare their heights

We need to find the ratio of their heights (hb / ha), to determine by how much higher Bob jumps compared to Adam: hb / ha = ((Vb^2) / (2g)) / ((Va^2) / (2g)) The '2g' in both the numerator and the denominator cancels out, and we can substitute Va = 0.5 * Vb: hb / ha = (Vb^2) / ((0.5 * Vb)^2)
05

5. Simplify the expression

When simplifying the expression, we get: hb / ha = (Vb^2) / (0.25 * Vb^2) = 1/0.25 = 4 So, the answer is: e) Bob jumps four times as high as Adam.

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

The trajectory of an object is given by the equation $$ x(t)=(4.35 \mathrm{~m})+(25.9 \mathrm{~m} / \mathrm{s}) t-\left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} $$ a) For which time \(t\) is the displacement \(x(t)\) at its maximum? b) What is this maximum value?

An object starts from rest and has an acceleration given by \(a=B t^{2}-\frac{1}{2} C t,\) where \(B=2.0 \mathrm{~m} / \mathrm{s}^{4}\) and \(C=-4.0 \mathrm{~m} / \mathrm{s}^{3}\). a) What is the object's velocity after 5.0 s? b) How far has the object moved after \(t=5.0\) s?

A train traveling at \(40.0 \mathrm{~m} / \mathrm{s}\) is headed straight toward another train, which is at rest on the same track. The moving train decelerates at \(6.0 \mathrm{~m} / \mathrm{s}^{2},\) and the stationary train is \(100.0 \mathrm{~m}\) away. How far from the stationary train will the moving train be when it comes to a stop?

In 2005, Hurricane Rita hit several states in the southern United States. In the panic to escape her wrath, thousands of people tried to flee Houston, Texas by car. One car full of college students traveling to Tyler, Texas, 199 miles north of Houston, moved at an average speed of \(3.0 \mathrm{~m} / \mathrm{s}\) for one-fourth of the time, then at \(4.5 \mathrm{~m} / \mathrm{s}\) for another one-fourth of the time, and at \(6.0 \mathrm{~m} / \mathrm{s}\) for the remainder of the trip. a) How long did it take the students to reach their destination? b) Sketch a graph of position versus time for the trip.

A car is traveling due west at \(20.0 \mathrm{~m} / \mathrm{s}\). Find the velocity of the car after \(37.00 \mathrm{~s}\) if its constant acceleration is \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) due east. Assume the acceleration remains constant. a) \(17.0 \mathrm{~m} / \mathrm{s}\) west b) \(17.0 \mathrm{~m} / \mathrm{s}\) east c) \(23.0 \mathrm{~m} / \mathrm{s}\) west d) \(23.0 \mathrm{~m} / \mathrm{s}\) east e) \(11.0 \mathrm{~m} / \mathrm{s}\) south
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Nuclear Physics

Read Explanation

Electricity

Read Explanation

Fluids

Read Explanation

Electric Charge Field and Potential

Read Explanation

Electromagnetism

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.