/*! 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 72 A balloon is rising at \(10 \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 72

A balloon is rising at \(10 \mathrm{m} / \mathrm{s}\) when its passenger throws a ball straight up at \(12 \mathrm{m} / \mathrm{s}\) relative to the balloon. How much later does the passenger catch the ball?

Short Answer

Expert verified
It would take approximately 4.49 seconds for the ball to return to the passenger's hand.

Step by step solution

01

Determine the Initial Velocity of the Ball

The initial velocity of the ball is the relative velocity at which it is thrown plus the velocity of the balloon. This would give an initial velocity of \(10 \, m/s + 12 \, m/s = 22 \, m/s\). This is when the ball leaves the hand of the passenger upward relative to the ground.
02

Apply the Physics Equation of Motion

Considering the upward direction as positive, we can apply the second equation of motion, \(h = ut - 0.5gt^2\), where \(h\) is the height, \(u\) is the initial velocity, \(g\) is the acceleration due to gravity and \(t\) is the time. The height we are considering is 0 (the height at which the passenger catches the ball is the same as the height from which it was thrown). So our equation becomes \(0 = 22t - 0.5 \cdot 9.8t^2\).
03

Solve for Time

Rewrite the equation in the form of a quadratic equation (\(at^2 + bt + c = 0\)) and then apply the quadratic formula (\(t = [-b \pm \sqrt{(b^2 - 4ac)}] / 2a\)). This gives us the equation \(0 = -4.9t^2 + 22t\), which simplifies to \(t = [-(-22) \pm \sqrt{((-22)^2 - (4 \cdot -4.9 \cdot 0))} ] / (2 \cdot -4.9)\). We solve this equation to obtain two times (t), one is 0 (the time when the ball was thrown) and the other will be the time when the ball comes back to the hand of the passenger.
04

Discard the Irrelevant Solution

Solving the quadratic equation gives two solutions (0s and 4.49s). Discard the 0s solution as it is the time at which ball was thrown and we are interested in the time when the ball returns to the passenger.

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 ball is dropped from rest at a height \(h_{0}\) above the ground. At the same instant, a second ball is launched with speed \(v_{0}\) straight up from the ground, at a point directly below where the other ball is dropped. (a) Find a condition on \(v_{0}\) such that the two balls will collide in mid-air. (b) Find an expression for the height at which they collide.

You're a consultant on a movie set, and the producer wants a car to drop so that it crosses the camera's field of view in time At. The field of view has height \(h .\) Derive an expression for the height above the top of the field of view from which the car should be released.

A hockey puck moving at 32 m/s slams through a wall of snow \(35 \mathrm{cm}\) thick. It emerges moving at \(18 \mathrm{m} / \mathrm{s}\). Assuming constant acceleration, find (a) the time the puck spends in the snow and (b) the thickness of a snow wall that would stop the puck entirely.

ThrustSSC, the world's first supersonic car, accelerates from rest to \(1000 \mathrm{km} / \mathrm{h}\) in \(16 \mathrm{s}\). What's its acceleration?

Does a speedometer measure speed or velocity?
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Force

Read Explanation

Electricity

Read Explanation

Oscillations

Read Explanation

Astrophysics

Read Explanation

Geometrical and Physical Optics

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.