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Chapter 3: Problem 68

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 at \(9.5 \mathrm{m}\) above the ground. How much time elapses before the ballast bag hits the ground?

Short Answer

Expert verified
The ballast bag hits the ground after 1.73 seconds.

Step by step solution

01

Identify Known Values

We know the initial height of the ballast bag above the ground is \( h_0 = 9.5 \, \text{m} \). The initial velocity of the bag relative to the ground is the same as the upward velocity of the balloon, which is \( v_{0} = 3.0 \, \text{m/s} \). The acceleration due to gravity, acting downwards, is \( g = 9.8 \, \text{m/s}^2 \).
02

Establish the Equation

To find the time \( t \) it takes for the bag to hit the ground, we use the equation of motion for an object under constant acceleration:\[ h = h_0 + v_{0}t - \frac{1}{2}gt^2 \]where \( h = 0 \) when the bag hits the ground.
03

Substitute Known Values

Substitute the known values into the equation:\[0 = 9.5 + 3.0t - \frac{1}{2} \times 9.8 \times t^2\] This simplifies to:\[ 0 = 9.5 + 3.0t - 4.9t^2 \]
04

Rearrange to Form a Quadratic Equation

Rearrange the terms to form a standard quadratic equation:\[ 4.9t^2 - 3.0t - 9.5 = 0 \]
05

Solve the Quadratic Equation

Apply the quadratic formula to solve for \( t \):\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 4.9 \), \( b = -3.0 \), and \( c = -9.5 \).
06

Calculate the Solution

First, find the discriminant:\[ b^2 - 4ac = (-3.0)^2 - 4 \times 4.9 \times (-9.5) \]\[ = 9 + 186.2 = 195.2 \]Now calculate \( t \):\[ t = \frac{-(-3.0) \pm \sqrt{195.2}}{2 \times 4.9} \]\[ t = \frac{3.0 \pm \sqrt{195.2}}{9.8} \]Solutions for \( t \) are:\[ t_1 = \frac{3.0 + 13.97}{9.8} = 1.73 \, \text{s} \]\[ t_2 = \frac{3.0 - 13.97}{9.8} = -1.12 \, \text{s} \]Since time cannot be negative, the time for the bag to hit the ground is \( t = 1.73 \, \text{s} \).

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Key Concepts

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

Kinematics
Kinematics deals with the motion of objects without considering the causes of this motion. It involves key quantities like displacement, velocity, and acceleration, which describe how things move.
When analyzing projectile motion, like our ballast bag, it is essential to identify these kinematic variables:
  • Initial Velocity ( \( v_{0} \)): This is the speed at which the object is moving at the start, which in our case is the 3.0 m/s velocity of the balloon.
  • Acceleration ( \( a \)): This is the rate at which the velocity of the object changes. Typically for free-falling objects near the Earth鈥檚 surface, the acceleration due to gravity ( \( g \)) is -9.8 m/s虏, acting downwards.
  • Displacement ( \( \Delta h \)): The change in the object鈥檚 position, initially given by the height of 9.5 m from the ground.
Using these kinematic principles, we formulate equations of motion that allow us to predict how long it will take for an object to reach a certain point, like our ballast hitting the ground.
Understanding these fundamentals helps clear any confusion regarding the paths objects take while in motion.
Quadratic Equation
In our problem, the equation describing the motion of the ballast bag is a quadratic equation. A quadratic equation is a second-degree polynomial that can generally be written in the form:\[ ax^2 + bx + c = 0 \]Quadratic equations frequently arise in kinematics, especially when analyzing free-fall problems where objects are subject to constant acceleration. When dealing with such equations:
  • Identify coefficients: In our example, the equation is \( 4.9t^2 - 3.0t - 9.5 = 0 \), where \( a = 4.9 \), \( b = -3.0 \), and \( c = -9.5 \).
  • Use the quadratic formula: The quadratic formula helps us find the roots of the equation:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
  • Evaluate solutions: Since we're dealing with time, we choose the solution that makes physical sense鈥攑ositive time is possible!
Solving quadratic equations allows us to predict outcomes accurately in physics problems, as seen when calculating the time it took our bag to hit the ground.
Free Fall
The concept of free fall addresses objects moving under the influence of gravitational force only. In our scenario, once the ballast bag is released, it experiences free fall.
Key aspects of free fall include:
  • Initial Conditions: The object may have an initial velocity. For our problem, the bag starts with the upward velocity of the balloon, which is \( 3.0 \, \text{m/s} \).
  • Acceleration due to Gravity: All free-falling objects have the same acceleration, \( g = 9.8 \, \text{m/s}^2 \), regardless of their mass.
  • Impact on Equations: The equations of motion incorporate these conditions to predict behavior over time. With \( h = h_0 + v_{0}t - \frac{1}{2}gt^2 \), we accommodate for both initial velocity and gravity.
By understanding free fall, we can analyze any object's motion under gravity alone, helping students grasp real-world implications of principles like those seen in this hot-air balloon problem.

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Most popular questions from this chapter

A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is \(670 \mathrm{m} / \mathrm{s}\). The gun is pointed directly at the center of the bull's-eye, but the bullet strikes the target \(0.025 \mathrm{m}\) below the center. What is the horizontal distance between the end of the rifle and the bull's-eye?

The captain of a plane wishes to proceed due west. The cruising speed of the plane is \(245 \mathrm{m} / \mathrm{s}\) relative to the air. A weather report indicates that a \(38.0-\mathrm{m} / \mathrm{s}\) wind is blowing from the south to the north. In what direction, measured with respect to due west, should the pilot head the plane?

E A bird watcher meanders through the woods, walking \(0.50 \mathrm{km}\) due east, \(0.75 \mathrm{km}\) due south, and \(2.15 \mathrm{km}\) in a direction \(35.0^{\circ}\) north of west. The time required for this trip is \(2.50 \mathrm{h}\). Determine the magnitude and direction (relative to due west) of the bird watcher's (a) displacement and (b) average velocity. Use kilometers and hours for distance and time, respectively.

A jetliner can fly 6.00 hours on a full load of fuel. Without any wind it flies at a speed of \(2.40 \times 10^{2} \mathrm{m} / \mathrm{s} .\) The plane is to make a round-trip by heading due west for a certain distance, turning around, and then heading due east for the return trip. During the entire flight, however, the plane encounters a \(57.8-\mathrm{m} / \mathrm{s}\) wind from the jet stream, which blows from west to east. What is the maximum distance that the plane can travel due west and just be able to return home?

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.
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