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From her bedroom window a girl drops a water-filled balloon to the ground, \(6.0 \mathrm{m}\) below. If the balloon is released from rest, how long is it in the air?

Short Answer

Expert verified
The balloon is in the air for about 1.1 seconds.

Step by step solution

01

Identify the Problem Type

This is a physics problem involving the free fall of an object under gravity. The balloon is released from rest, so its initial velocity is zero. We need to find the time it takes for the balloon to fall to the ground.
02

Identify Given Values

The height from which the balloon is dropped is given as \(h = 6.0\, \text{m}\). The initial velocity \(v_0\) is \(0\, \text{m/s}\) because the balloon is released from rest. The acceleration \(a\) due to gravity is \(g = 9.8\, \text{m/s}^2\).
03

Use the Free Fall Equation

To find the time \(t\), we use the equation of motion for free fall: \(h = \frac{1}{2} g t^2\). This equation models the distance fallen as a function of time and gravitational acceleration.
04

Rearrange the Equation to Solve for Time

Rearrange the equation \(h = \frac{1}{2} g t^2\) to solve for \(t\):\[t = \sqrt{\frac{2h}{g}}.\]
05

Substitute the Known Values

Substitute \(h = 6.0\, \text{m}\) and \(g = 9.8\, \text{m/s}^2\) into the equation:\[t = \sqrt{\frac{2 \times 6.0}{9.8}}.\]
06

Calculate the Time in Air

Calculate \(t\) using the substituted values:\[t = \sqrt{\frac{12}{9.8}} \approx \sqrt{1.22449} \approx 1.1\, \text{s}.\]The time the balloon is in the air is approximately \(1.1\, \text{s}\).

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Key Concepts

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

Acceleration Due to Gravity
Gravity is the force that pulls objects toward the Earth, and it is represented by the symbol "g." On Earth, the standard acceleration due to gravity is approximately 9.8 meters per second squared, denoted as \(g = 9.8 \, \text{m/s}^2\). This means that any object in free fall will accelerate downwards at this constant rate, assuming there is no air resistance.
The role of gravity in free fall is crucial. It determines how fast an object initially at rest will speed up as it falls. For every second an object is in free fall, its speed increases by approximately 9.8 meters per second.
Gravity acts equally on all objects, regardless of their mass. So, whether a feather or a rock is dropped, they both fall at the same rate absent air resistance, demonstrating the universal nature of gravitational acceleration.
Equations of Motion
Equations of motion form the backbone of physics that describes how objects move. In a free fall scenario, we primarily use one equation: \[ h = \frac{1}{2} g t^2 \]This specific formula tells us how distance (or height) relates to time when an object falls freely under gravity's influence.
Each part of the equation has importance:
  • \(h\) is the height or distance the object falls.
  • \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\) on Earth).
  • \(t\) is the time in seconds the object is in motion.
Using these equations, we can derive other quantities such as the final velocity and time of flight, assuming we know the initial conditions. The amazing part is how these relationships are consistently reliable in describing many different motion scenarios.
Initial Velocity
Initial velocity is a term that describes the speed and direction of an object before any forces act upon it. In many free fall scenarios, such as the one with the balloon, the object is "dropped." This means it is released from rest, and its initial velocity \((v_0)\) is 0 meters per second.
In problems dealing with free fall, starting with an initial velocity of zero simplifies the equations of motion. The absence of upward or forward motion ensures that the only influencing force is gravity, which constantly changes the velocity over time as the object gains speed while falling.
Understanding the initial state of an object’s motion is crucial when solving physics problems. It allows us to calculate eventual speed, total distance traveled, and time taken using simplified forms of motion equations since additional variables due to initial motion are absent.

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Most popular questions from this chapter

The Kentucky Derby is held at the Churchill Downs track in Louisville, Kentucky. The track is one and one-quarter miles in length. One of the most famous horses to win this event was Secretariat. In 1973 he set a Derby record that would be hard to beat. His average acceleration during the last four quarter-miles of the race was \(+0.0105 \mathrm{m} / \mathrm{s}^{2}\). His velocity at the start of the final mile \((x=+1609 \mathrm{m})\) was about \(+16.58 \mathrm{m} / \mathrm{s} .\) The acceleration, although small, was very important to his victory. To assess its effect, determine the difference between the time he would have taken to run the final mile at a constant velocity of \(+16.58 \mathrm{m} / \mathrm{s}\) and the time he actually took. Although the track is oval in shape, assume it is straight for the purpose of this problem.

An Australian emu is running due north in a straight line at a speed of \(13.0 \mathrm{m} / \mathrm{s}\) and slows down to a speed of \(10.6 \mathrm{m} / \mathrm{s}\) in \(4.0 \mathrm{s}\). (a) What is the direction of the bird's acceleration? (b) Assuming that the acceleration remains the same, what is the bird's velocity after an additional \(2.0 \mathrm{s}\) has elapsed?

A motorcycle has a constant acceleration of \(2.5 \mathrm{m} / \mathrm{s}^{2} .\) Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from (a) 21 to \(31 \mathrm{m} / \mathrm{s},\) and (b) 51 to \(61 \mathrm{m} / \mathrm{s} ?\)

The initial velocity and acceleration of four moving objects at a given instant in time are given in the following table. Determine the final speed of each of the objects, assuming that the time elapsed since \(t=0\) s is 2.0 s. $$ \begin{array}{lcc} & \text { Initial velocity } v_{0} & \text { Acceleration } a \\\\\hline \text { (a) } & +12 \mathrm{m} / \mathrm{s} & +3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (b) } & +12 \mathrm{m} / \mathrm{s} & -3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (c) } & -12 \mathrm{m} / \mathrm{s} & +3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (d) } & -12 \mathrm{m} / \mathrm{s} & -3.0 \mathrm{m} / \mathrm{s}^{2} \\ \hline\end{array}$$

A cheetah is hunting. Its prey runs for \(3.0 \mathrm{s}\) at a constant velocity of \(+9.0 \mathrm{m} / \mathrm{s} .\) Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time?

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