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Chapter 2: Problem 50

A hot-air balloon is ascending at the rate of \(12 \mathrm{~m} / \mathrm{s}\) and is \(80 \mathrm{~m}\) above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?

Short Answer

Expert verified
(a) Approximately 6.4 seconds, (b) Around 50.9 m/s.

Step by step solution

01

Calculate Initial Conditions

The package has the same initial upwards velocity as the balloon. Initial velocity \(u = 12 \text{ m/s}\) and initial height \(h = 80 \text{ m}\) above the ground.
02

Use Kinematic Equation to Determine Time

We need to find the time \(t\) it takes for the package to reach the ground. The kinematic equation for vertical motion under gravity is given by \(h = ut - \frac{1}{2}gt^2\) where \(g\) is the acceleration due to gravity, \(9.8 \text{ m/s}^2\). Setting \(h = -80\) (since we take upwards as positive), solve for \(t\): \(-80 = 12t - \frac{1}{2}(9.8)t^2\). This simplifies to \(4.9t^2 - 12t - 80 = 0\).
03

Solve the Quadratic Equation

Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4.9\), \(b = -12\), and \(c = -80\). Substitute these values in to find \(t\). Calculate the discriminant \(b^2 - 4ac = 12^2 - 4 \times 4.9 \times (-80)\). Compute the roots to find the positive time value.
04

Determine Impact Speed Using Final Velocity

The final speed is found using the kinematic equation \(v = u + gt\). Substitute \(u = 12 \text{ m/s}\), \(g = 9.8 \text{ m/s}^2\), and the time from Step 3. Compute \(v\): \(v = 12 + 9.8t\). Use the value of \(t\) to find \(v\).

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Key Concepts

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

Projectile Motion
Understanding projectile motion can initially seem challenging, but it becomes simpler when you break it down. In projectile motion, an object is thrown into space and is subject to gravity, influencing its trajectory. The movement can be separated into two components:
  • Horizontal motion鈥攃onstant velocity since there is no horizontal acceleration in the absence of air resistance.
  • Vertical motion鈥攁ffected by the acceleration due to gravity, which acts downward.
In our exercise, even though the package is initially moving upwards, it forms a part of projectile motion as it is in free flight after being dropped. The motion is affected by two factors: the initial velocity upwards and the acceleration due to gravity pulling it downwards, creating a parabolic trajectory until it reaches the ground.
Free Fall
Free fall is a specific type of motion where gravity is the only force acting on an object. Once the package is released from the balloon, it enters free fall because it acts under the influence of gravity alone without any additional forces in the vertical direction. Despite having an initial upwards velocity, it鈥檚 important to treat its journey as a free fall in terms of understanding.
In this scenario, the package initially travels upwards, slows down, and then accelerates downward. We use the kinematic equation: \[ h = ut - \frac{1}{2}gt^2 \]where:
  • \( h \) is the vertical distance (negative in our example, as the package descends),
  • \( u \) is the initial velocity upwards,
  • \( g \) is the acceleration due to gravity, typically \( 9.8 \, \text{m/s}^2 \),
  • \( t \) is the time taken to fall.
By substituting the values and seeking solutions to the quadratic equation, the time to reach the ground is determined.
Acceleration due to Gravity
A fundamental concept in projectile motion and free fall is the acceleration due to gravity, denoted as \( g \). It is crucial for calculating the behavior of objects in motion near the Earth's surface where \( g \) is approximately \( 9.8 \, \text{m/s}^2 \).
This constant acceleration impacts the package immediately after it is released. Regardless of the direction it was initially moving, gravity will eventually pull it toward the ground. The constant negative acceleration causes the upward velocity to decrease until it halts, after which the package gains speed in the downward direction.
In the problem, this acceleration plays a key role in finding both the time the package takes to hit the ground and its final speed, calculated using:\[ v = u + gt \]where:
  • \( v \) is the final velocity,
  • \( u \) is the initial upward velocity,
  • \( t \) is the total time in free fall.
Using these principles helps in accurately predicting the dynamics of objects in free fall, which is essential in physics applications.

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Most popular questions from this chapter

In a particle accelerator, an electron enters a region in which it accelerates uniformly in a straight line from a speed of \(4.00 \times 10^{5}\) \(\mathrm{m} / \mathrm{s}\) to a speed of \(6.00 \times 10^{7} \mathrm{~m} / \mathrm{s}\) in a distance of \(3.00 \mathrm{~cm}\). For what time interval does the electron accelerate?

(a) If the maximum acceleration that is tolerable for passengers in a subway train is \(1.34 \mathrm{~m} / \mathrm{s}^{2}\) and subway stations are located \(880 \mathrm{~m}\) apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for \(20 \mathrm{~s}\) at each station, what is the maximum average speed of the train, from one start-up to the next? (d) Graph \(x, v\), and \(a\) versus \(t\) for the interval from one start-up to the next.

An electron, starting from rest and moving with a constant acceleration, travels \(2.00 \mathrm{~cm}\) in \(5.00 \mathrm{~ms}\). What is the magnitude of this acceleration?

The position of an object moving along an \(x\) axis is given by \(x=3 t-4 t^{2}+t^{3}\), where \(x\) is in meters and \(t\) in seconds. Find the position of the object at the following values of \(t\) : (a) \(1 \mathrm{~s}\), (b) \(2 \mathrm{~s}\), (c) \(3 \mathrm{~s}\), and (d) \(4 \mathrm{~s}\). (e) What is the object's displacement between \(t=0\) and \(t=4 \mathrm{~s} ?\) (f) What is its average velocity for the time interval from \(t=2 \mathrm{~s}\) to \(t=4 \mathrm{~s} ?(\mathrm{~g})\) Graph \(x\) versus \(t\) for \(0 \leq t \leq 4 \mathrm{~s}\) and indicate how the answer for (f) can be found on the graph.

A rock is thrown downward from an unknown height above the ground with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\). It strikes the ground \(3.0 \mathrm{~s}\) later. Determine the initial height of the rock above the ground.
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