/*! 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 17 A hot-air balloon is rising stra... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 17

A hot-air balloon is rising straight up with a speed of \(3.0 \mathrm{~m} / \mathrm{s}\). A ballast bag is released from rest relative to the balloon when it is \(9.5 \mathrm{~m}\) above the ground. How much time elapses before the ballast bag hits the ground?

Short Answer

Expert verified
The ballast bag hits the ground in approximately 1.22 seconds.

Step by step solution

01

Understanding the problem

The balloon is moving upward with an initial velocity of \( 3.0 \mathrm{~m/s} \) and the ballast bag is released from rest relative to the balloon at a height of \( 9.5 \mathrm{~m} \). The goal is to find the time it takes for the ballast bag to hit the ground after being released.
02

Recognizing initial conditions

When the ballast bag is released, it has the same upward velocity as the balloon, \( v_0 = 3.0 \mathrm{~m/s} \). The height from which it is released is \( h = 9.5 \mathrm{~m} \). The acceleration due to gravity will act downward on the bag, \( g = 9.81 \mathrm{~m/s^2} \).
03

Applying the kinematic equation

Use the kinematic equation for an object in free-fall:\[ h = v_0 t + \frac{1}{2} a t^2\]Since the displacement will be negative (as it falls to the ground), the equation becomes:\[ -9.5 = 3.0t - \frac{1}{2} \times 9.81 t^2 \]
04

Solving the quadratic equation

Rearrange the equation to standard quadratic form:\[ 0 = -4.905t^2 + 3.0t - 9.5 \]Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -4.905 \), \( b = 3.0 \), and \( c = -9.5 \), solve for \( t \).
05

Calculating time values

Calculate the discriminant: \[ b^2 - 4ac = 3.0^2 - 4 \times (-4.905) \times (-9.5) = 9 - 186.69 = -177.69 \]So, there was an error in rearranging or discriminant calculation. Let's go over the steps again.
06

Review of calculation

Let's correct the mistake in solving:Calculate the discriminant again properly using correct formula:\[ b^2 - 4ac = 3.0^2 - 4 \times (-4.905) \times (-9.5)\]This becomes:\[ 9 + 186.69 = 195.69\]Plug into the quadratic formula:\[ t = \frac{-3.0 \pm \sqrt{195.69}}{-9.81} \]Calculate the positive root to find the time.
07

Selecting the positive root

Find the roots from the quadratic formula:\[ t = \frac{-3.0 + 13.992}{-9.81} \approx 1.22 \text{ seconds}\]The negative root is physically meaningless as time cannot be negative.

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!

Key Concepts

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

Kinematic Equations
In physics, kinematic equations describe the motion of objects. They allow us to calculate unknown variables, such as time, displacement, or velocity.
These equations are particularly useful for objects in free-fall because they incorporate aspects like initial velocity and acceleration. In our exercise, we use the equation:
  • \( h = v_0 t + \frac{1}{2} a t^2 \)

Here, \( h \) represents displacement, \( v_0 \) is the initial velocity, \( a \) is acceleration, and \( t \) is time. This equation calculates how far an object travels over time when subjected to constant acceleration.
For the ballast bag, the initial velocity \( v_0 \) matches the balloon's upward speed (3.0 m/s). The acceleration \( a \) is due to gravity, which changes the motion dynamics.
In this problem, knowing the initial height (9.5 m) and rearranging our equation, we find how long it takes for the bag to reach the ground.
Quadratic Formula
The quadratic formula is essential for finding solutions to quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). This formula is particularly useful when solving kinematic equations that describe nonlinear paths.
The formula is:
  • \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In the context of our exercise, the equation \(-4.905t^2 + 3.0t - 9.5 = 0\) is quadratic. The coefficients are \( a = -4.905 \), \( b = 3.0 \), and \( c = -9.5 \). We apply the quadratic formula to determine the time \( t \), which gives us when the ballast will hit the ground.
We make sure to select the positive root from the quadratic equation, as time can't be negative, giving us around 1.22 seconds for the bag to land.
Acceleration Due to Gravity
Acceleration due to gravity is fundamental in studying free-fall motion. It represents the constant rate at which objects accelerate towards Earth if only under gravitational force.
Its average value on Earth is \(-9.81 \, \text{m/s}^2\). This means every second, an object's velocity downwards increases by approximately \(9.81 \, \text{m/s}\).
For the ballast bag problem, this acceleration counteracts the initial upwards velocity, eventually bringing the bag back down to the ground.
When mentioning acceleration in kinematic equations:
  • A negative sign indicates a downward force or deceleration, relative to an initial upward motion.

Understanding this constant allows us to predict how quickly and where an object will move during free-fall, making calculations like ours feasible.

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

Two friends, Barbara and Neil, are out rollerblading. With respect to the ground, Barbara is skating due south. Neil is in front of her and to her left. With respect to the ground, he is skating due west. (a) Does Barbara see him moving toward the east or toward the west? (b) Does Barbara see him moving toward the north or toward the south? (c) Considering your answers to parts (a) and (b), how does Barbara see Neil moving relative to herself, toward the east and north, toward the east and south, toward the west and north, or toward the west and south? Justify your answers in each case. With respect to the ground, Barbara is skating due south at a speed of \(4.0 \mathrm{~m} / \mathrm{s}\). With respect to the ground, Neil is skating due west at a speed of \(3.2 \mathrm{~m} / \mathrm{s}\). Find Neil's velocity (magnitude and direction relative to due west) as seen by Barbara. Make sure that your answer agrees with your answer to part (c) of the Concept Questions.

An airplane with a speed of \(97.5 \mathrm{~m} / \mathrm{s}\) is climbing upward at an angle of \(50.0^{\circ}\) with respect to the horizontal. When the plane's altitude is \(732 \mathrm{~m}\), the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

Two trees have perfectly straight trunks and are both growing perpendicular to the flat horizontal ground beneath them. The sides of the trunks that face each other are separated by \(1.3 \mathrm{~m}\). A frisky squirrel makes three jumps in rapid succession. First, he leaps from the foot of one tree to a spot that is \(1.0 \mathrm{~m}\) above the ground on the other tree. Then, he jumps back to the first tree, landing on it at a spot that is \(1.7 \mathrm{~m}\) above the ground. Finally, he leaps back to the other tree, now landing at a spot that is \(2.5 \mathrm{~m}\) above the ground. What is the magnitude of the squirrel's displacement?

A puck is moving on an air hockey table. Relative to an \(x, y\) coordinate system at time \(t=0 \mathrm{~s},\) the \(x\) components of the puck's initial velocity and acceleration are \(v_{0 x}=+1.0 \mathrm{~m} / \mathrm{s}\) and \(a_{x}=+2.0 \mathrm{~m} / \mathrm{s}^{2} .\) The \(y\) components of the puck's initial velocity and acceleration are \(v_{0 y}=+2.0 \mathrm{~m} / \mathrm{s}\) and \(a_{y}=-2.0 \mathrm{~m} / \mathrm{s}^{2}\) Is the magnitude of the \(x\) component of the velocity increasing or decreasing in time? Is the magnitude of the \(y\) component of the velocity increasing or decreasing in time? Find the magnitude and direction of the puck's velocity at a time of \(t=0.50 \mathrm{~s}\). Specify the direction relative to the \(+x\) axis. Be sure that your calculations are consistent with your answers to the Concept Questions.

On a pleasure cruise a boat is traveling relative to the water at a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) due south. Relative to the boat, a passenger walks toward the back of the boat at a speed of \(1.5 \mathrm{~m} / \mathrm{s} .\) (a) What is the magnitude and direction of the passenger's velocity relative to the water? (b) How long does it take for the passenger to walk a distance of \(27 \mathrm{~m}\) on the boat? (c) How long does it take for the passenger to cover a distance of \(27 \mathrm{~m}\) on the water?
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Geometrical and Physical Optics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Kinematics Physics

Read Explanation

Nuclear Physics

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.