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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 approximately 1.11 seconds.

Step by step solution

01

Identify Known Values

The problem states that the balloon falls a distance of 6.0 meters. It is released from rest, so the initial velocity \( u = 0 \) m/s. The acceleration due to gravity \( g = 9.8 \) m/s².
02

Select the Appropriate Formula

To find the time \( t \) the balloon is in the air, we'll use the kinematic equation: \[ s = ut + \frac{1}{2}gt^2 \] Where \( s \) is the distance fallen, \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( t \) is the time.
03

Substitute Known Values

Substitute \( s = 6.0 \) m, \( u = 0 \) m/s, and \( g = 9.8 \) m/s² into the equation: \[ 6.0 = 0 \, t + \frac{1}{2} \times 9.8 \times t^2 \] This simplifies to: \[ 6.0 = 4.9t^2 \]
04

Solve for Time \( t \)

Rearrange the equation to find \( t^2 \): \[ t^2 = \frac{6.0}{4.9} \] Calculate \( t \): \[ t = \sqrt{\frac{6.0}{4.9}} \] Calculate the square root: \[ t \approx 1.11 \] seconds.

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Key Concepts

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

Understanding Free Fall
Free fall is a fascinating concept in physics, referring to the motion of objects that are falling solely under the influence of gravity. Imagine letting go of a water balloon and watching it plummet to the ground. That drop is a perfect example of free fall. In this scenario, no other forces, like air resistance, are acting significantly on the balloon.

During free fall, an object experiences a constant acceleration towards the Earth, known as the acceleration due to gravity. What's intriguing here is that regardless of the object's mass, the rate at which it falls stays consistent. This means a heavy rock and a feather (in a vacuum) fall at the same rate!

Key points about free fall include:
  • Initial velocity can often be zero, especially if the fall starts from a rest position.
  • The only force acting is gravity, simplifying calculations.
  • Gravity ensures a uniform acceleration downwards.
Understanding free fall helps in grasping more complex physics motions and real-world applications.
Kinematic Equations in Action
Kinematic equations are essential tools in physics used to describe motion quantitatively. In our balloon example, we needed to find out how long it took for the balloon to hit the ground. To solve this, kinematic equations are used, especially ideal for scenarios involving constant accelerations, such as free fall.

One of the most useful kinematic equations is the one that relates distance, initial velocity, time, and acceleration: \[ s = ut + \frac{1}{2}gt^2 \]

This equation helped determine how long the balloon was in the air by considering:
  • Initial velocity \( u \), which was zero since it was dropped.
  • The acceleration due to gravity \( g \), which affected its motion.
  • The distance \( s \) the balloon traveled.
Kinematic equations effectively decode the mysteries of motion, providing means to predict and analyze the movement of objects. This exploration empowers us to connect equations with real-life scenarios seamlessly.
Exploring Acceleration Due to Gravity
The acceleration due to gravity (denoted as \( g \)) is a crucial constant in physics, particularly when studying free fall scenarios. This constant represents the rate at which objects accelerate towards Earth when dropped. Typically, on the surface of the Earth, this value is approximately \( 9.8 \) m/s².

Understanding this constant is pivotal because:
  • It applies universally to all objects in free fall near the Earth's surface.
  • It allows for precise calculations of time, velocity, and displacement during free fall.
  • It simplifies complex motion to basics, aiding in clearer problem-solving strategies.


In our example, this consistent value enabled us to predict the balloon's motion precisely. With \( g = 9.8 \) m/s², we accurately calculated the time taken for the balloon to hit the ground using kinematic equations. This fundamental knowledge not only supports theoretical learning but also has real-world applications, such as calculating trajectories and understanding where objects will land.

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Most popular questions from this chapter

A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position \(x=0 \mathrm{m}\), with an initial velocity of \(+5.0 \mathrm{m} / \mathrm{s}\) and a constant acceleration due to the fan. The direction to the right is positive. The cart reaches a maximum position of \(x=+12.5 \mathrm{m},\) where it begins to travel in the negative direction. Find the acceleration of the cart.

The three-toed sloth is the slowest-moving land mammal. On the ground, the sloth moves at an average speed of \(0.037 \mathrm{m} / \mathrm{s}\), considerably slower than the giant tortoise, which walks at \(0.076 \mathrm{m} / \mathrm{s}\). After 12 minutes of walking, how much further would the tortoise have gone relative to the sloth?

Starting at \(x=-16 \mathrm{m}\) at time \(t=0 \mathrm{s},\) an object takes \(18 \mathrm{s}\) to travel \(48 \mathrm{m}\) in the \(+x\) direction at a constant velocity. Make a position-time graph of the object's motion and calculate its velocity.

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is \(6.00 \mathrm{m}\). The stones are thrown with the same speed of \(9.00 \mathrm{m} / \mathrm{s}\). Find the location (above the base of the cliff) of the point where the stones cross paths.

A diver springs upward with an initial speed of \(1.8 \mathrm{m} / \mathrm{s}\) from a \(3.0-\mathrm{m}\) board. (a) Find the velocity with which he strikes the water. / Hint: When the diver reaches the water, his displacement is \(y=-3.0 \mathrm{m}\) (measured from the board), assuming that the dowmward direction is chosen as the negative direction./ (b) What is the highest point he reaches above the water?

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