/*! 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 129 Suppose the balloon is descendin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 129

Suppose the balloon is descending with a constant speed of \(4.2 \mathrm{m} / \mathrm{s}\) when the bag of sand comes loose at a height of \(35 \mathrm{m}\). (a) How long is the bag in the air? (b) What is the speed of the bag when it is \(15 \mathrm{m}\) above the ground?

Short Answer

Expert verified
(a) The bag is in the air for approximately 2.54 seconds. (b) Its speed is 20.23 m/s at 15 meters above the ground.

Step by step solution

01

Identifying Given Values and Known Equations

We know that the balloon is descending with a constant speed of 4.2 m/s, and the bag of sand is initially at a height of 35 m above the ground. We need to calculate the time the bag is in the air (part a) and its speed when it is 15 m above the ground (part b). The primary equation for free fall is the kinematic equation: \[ h = v_i t + \frac{1}{2}gt^2 \]where \( h \) is the height, \( v_i \) is the initial velocity, \( g \) is the acceleration due to gravity (approximated as 9.8 m/s虏), and \( t \) is the time.
02

Solving for Time in the Air

The bag has an initial downward velocity \( v_i = 4.2 \ \mathrm{m/s} \). We use the kinematic equation: \[ 0 = 35 + 4.2t + \frac{1}{2}(-9.8)t^2 \]This is a quadratic equation. Rearranging gives:\[ 4.9t^2 + 4.2t - 35 = 0 \]We use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4.9 \), \( b = 4.2 \), and \( c = -35 \). Calculating the discriminant and solving yields two possible times. Since time cannot be negative, we use the positive value.
03

Calculating Discriminant and Solving Quadratic Equation

The discriminant is given by:\[ b^2 - 4ac = (4.2)^2 - 4(4.9)(-35) \]Calculating gives:\[ 17.64 + 686 = 703.64 \]Taking the square root and solving the quadratic equation, we get:\[ t = \frac{-4.2 \pm \sqrt{703.64}}{9.8} \]One of the solutions is \( t \approx 2.54 \) seconds (positive value). The negative root does not make sense in this context.
04

Finding Speed at 15 Meters Above Ground

When the bag is 15 meters from the ground, we know it has fallen 20 meters. We use the kinematic equation for velocity, \[ v^2 = v_i^2 + 2g h \] where \( h = 20 \ \mathrm{m} \) from \( 35 \ \mathrm{m} \) to \( 15 \ \mathrm{m} \). Plug in values:\[ v^2 = 4.2^2 + 2(9.8)(20) \]\[ v^2 = 17.64 + 392 = 409.64 \]Therefore, \( v \approx \sqrt{409.64} = 20.23 \ \mathrm{m/s} \).

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
Kinematic equations describe the motion of objects and are particularly useful in scenarios involving constant acceleration, such as free fall. In this exercise, we use one of the kinematic equations for determining the time a sandbag takes to hit the ground as well as its speed at a certain height. The equation in use is:\[ h = v_i t + \frac{1}{2} g t^2 \]Where:
  • h is the displacement or height,
  • vi is the initial velocity,
  • g is the acceleration due to gravity,
  • t is the time in seconds.

For the sandbag, the kinematic equations help calculate how long it is in the air and at what speed it reaches a point before touching the ground. These equations consider the uniform motion given by gravity. This exercise focuses on adapting them to a descending object with an initial velocity.
Quadratic Equation
The quadratic equation plays a vital role when solving problems involving kinematic equations where time is unknown. In our scenario, substituting the known values into the kinematic equation results in a quadratic form:\[ 4.9 t^2 + 4.2 t - 35 = 0 \]Here, we transform the kinematic equation into a typical quadratic \( ax^2 + bx + c = 0 \).
  • The coefficients are: a = 4.9, b = 4.2, and c = -35.

The discriminant \( b^2 - 4ac \) helps determine the nature of the quadratic roots:
  • If positive, real and distinct roots exist, helpful for solving time in physics problems.
  • Here, we use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find time, ensuring realistic, positive values applicable to time.
Acceleration Due to Gravity
The free fall motion is significantly influenced by gravity, a constant force that accelerates objects at approximately 9.8 m/s虏 towards the Earth. This uniform acceleration ensures that, regardless of the object's initial velocity, gravity will modify its speed over time, as demonstrated in the exercise.
In this problem, gravity is crucial for determining two things:
  • How quickly the sandbag accelerates as it falls.
  • Changes in velocity over the distance, using equations like \( v^2 = v_i^2 + 2gh \).

Importantly, gravity affects motion uniformly, establishing a predictable acceleration pattern crucial to solving such physics problems.
Initial Velocity
Initial velocity refers to the speed and direction an object has when it starts its journey. For the sandbag in this exercise, it begins with an initial downward velocity of 4.2 m/s, same as the descending speed of the balloon from which it detached.
  • The initial velocity is vital because it combines with gravity鈥檚 effect to determine the bag's future velocity through the air.
  • It also sets up the starting conditions for the kinematic equations we use, affecting both time in the air and speed changes.

Without considering the initial velocity, our calculations would underestimate how natural forces alter its falling motion, leading to incorrectly predicting the time it remains airborne and its velocity at a specific height.

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 infamous chicken is dashing toward home plate with a speed of \(5.8 \mathrm{m} / \mathrm{s}\) when he decides to hit the dirt. The chicken slides for 1.1 s, just reaching the plate as he stops (safe, of course). (a) What are the magnitude and direction of the chicken's acceleration? (b) How far did the chicken slide?

A youngster bounces straight up and down on a trampoline. Suppose she doubles her initial speed from \(2.0 \mathrm{m} / \mathrm{s}\) to \(4.0 \mathrm{m} / \mathrm{s} .\) (a) By what factor does her time in the air increase? (b) By what factor does her maximum height increase? (c) Verify your answers to parts (a) and (b) with an explicit calculation.

The world's highest fountain of water is located, appropriately enough, in Fountain Hills, Arizona. The fountain rises to a height of \(560 \mathrm{ft}(5\) feet higher than the Washington Monument). (a) What is the initial speed of the water? (b) How long does it take for water to reach the top of the fountain?

In a physics lab, students measure the time it takes a small cart to slide a distance of \(1.00 \mathrm{m}\) on a smooth track inclined at an angle \(\theta\) above the horizontal. Their results are given in the following table. $$ \begin{array}{llll} \hline \theta & 10.0^{\circ} & 20.0^{\circ} & 30.0^{\circ} \\ \text { time, } \mathrm{s} & 1.08 & 0.770 & 0.640 \\ \hline \end{array} $$

An astronaut on the Moon drops a rock straight downward from a height of \(1.25 \mathrm{m}\). If the acceleration of gravity on the Moon is \(1.62 \mathrm{m} / \mathrm{s}^{2},\) what is the speed of the rock just before it lands?
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Physics of Motion

Read Explanation

Classical Mechanics

Read Explanation

Force

Read Explanation

Electricity

Read Explanation

Dynamics

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.