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Chapter 2: Problem 49

A hot-air balloon is rising upward with a constant speed of\(2.50 \mathrm{m} / \mathrm{s} .\) When the balloon is \(3.00 \mathrm{m}\) above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?

Short Answer

Expert verified
0.57 seconds

Step by step solution

01

Understand the problem

We need to find the time it takes for the compass, initially at a height of 3.00 m and rising at 2.50 m/s, to hit the ground after being dropped. When it is released, it will have an initial velocity of 2.50 m/s and it will be affected by the acceleration due to gravity.
02

Define the initial conditions

The initial velocity of the compass (\(v_0\)) is 2.50 m/s (upward), and the initial height (\(h_0\)) is 3.00 m. The final position when it hits the ground is 0 m. The acceleration due to gravity (\(g\)) is -9.81 m/s虏 (since it acts downward).
03

Set up the kinematic equation

Use the kinematic equation for position:\[ h = h_0 + v_0 t - \frac{1}{2} g t^2 \]Set \(h = 0\) (ground level), \(h_0 = 3.00\) m, \(v_0 = 2.50\) m/s, and \(g = 9.81\) m/s虏. Substitute these values into the equation to get:\[ 0 = 3.00 + 2.50 t - \frac{1}{2} (9.81) t^2 \]
04

Simplify the equation

Rearrange the equation:\[ 0 = 3.00 + 2.50t - 4.905t^2 \]Reorder it into the form \(at^2 + bt + c = 0\):\[ -4.905t^2 + 2.50t + 3.00 = 0 \]
05

Solve the quadratic equation

Use the quadratic formula\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = -4.905\), \(b = 2.50\), and \(c = 3.00\). Calculate the discriminant:\[ b^2 - 4ac = (2.50)^2 - 4(-4.905)(3.00) = 6.25 + 58.86 = 65.11 \]Substitute back to find \(t\):\[ t = \frac{-2.50 \pm \sqrt{65.11}}{-9.81} \]
06

Calculate the time

Calculate the square root and solve for \(t\):\[ \sqrt{65.11} \approx 8.07 \]\[ t = \frac{-2.50 + 8.07}{9.81} \approx \frac{5.57}{9.81} \approx 0.57 \text{ seconds} \](The negative solution will not be physically meaningful in this context.)

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Key Concepts

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

Gravity
Gravity is a fundamental force that attracts two masses towards each other. On Earth, gravity gives weight to physical objects and causes them to fall towards the ground when dropped. In kinematics, which is the study of motion, gravity is a constant force that affects the movement of objects.- **Direction:** Gravity acts downwards, towards the center of the Earth.- **Magnitude:** On Earth, it has a standard acceleration of \(9.81 \text{ m/s}^2\).When the compass is dropped from the hot-air balloon, gravity pulls it downward, which affects how fast and how far it falls. Understanding gravity is essential to calculate the motion of the compass after being released.
Initial Velocity
Initial velocity refers to the speed and direction an object has when it starts its motion. Whenever an object is released or set into motion, it has an initial speed that must be considered.For the compass in the exercise:- The initial velocity is \(2.50 \text{ m/s}\) upward because it starts moving with the balloon.- This initial velocity affects the time it takes for the compass to hit the ground because it initially moves upward before gravity pulls it down.Understanding the concept of initial velocity is crucial in kinematic equations where the motion needs to be calculated.
Quadratic Equation
A quadratic equation is a polynomial equation of the second degree, typically in the form:\[ ax^2 + bx + c = 0 \]In kinematics, quadratic equations are often used to calculate the time it takes for an object to reach a certain position, like when determining when the compass hits the ground.Important points about quadratic equations:- Solving them often requires rearranging the equation properly.- The solutions can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this exercise, the equation derived from the motion of the compass was rearranged into a quadratic form to solve for the time at which the compass reaches the ground.
Acceleration Due to Gravity
Acceleration due to gravity is the rate at which an object increases its velocity when falling freely towards Earth. Denoted by \(g\), this value is approximately \(9.81 \text{ m/s}^2\), indicating that an object's velocity increases by 9.81 meters per second every second that it is in free fall.Important aspects of acceleration due to gravity in this context include:- It's always directed downwards towards the Earth's center.- It affects all objects equally regardless of their masses.In the scenario of the dropped compass, the acceleration due to gravity slows down the upward motion (due to its initial velocity) and then increases its speed downward until it hits the ground. Understanding this concept is essential for computing the motion using kinematic equations.

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Most popular questions from this chapter

A police car is traveling at a velocity of \(18.0 \mathrm{m} / \mathrm{s}\) due north, when a car zooms by at a constant velocity of \(42.0 \mathrm{m} / \mathrm{s}\) due north. After a reaction time of 0.800 s the policeman begins to pursue the speeder with an acceleration of \(5.00 \mathrm{m} / \mathrm{s}^{2} .\) Including the reaction time, how long does it take for the police car to catch up with the speeder?

The space shuttle travels at a speed of about \(7.6 \times 10^{3} \mathrm{m} / \mathrm{s} .\) The blink of an astronaut's eye lasts about \(110 \mathrm{ms}\). How many football fields (length \(=91.4 \mathrm{m})\) does the shuttle cover in the blink of an eye?

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?

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 motorcycles are traveling due east with different velocities. However, four seconds later, they have the same velocity. During this foursecond interval, cycle A has an average acceleration of \(2.0 \mathrm{m} / \mathrm{s}^{2}\) due east, while cycle B has an average acceleration of \(4.0 \mathrm{m} / \mathrm{s}^{2}\) due east. By how much did the speeds differ at the beginning of the four-second interval, and which motorcycle was moving faster?.
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