/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 90 A hot-air balloon is descending ... [FREE SOLUTION] | 91Ó°ÊÓ

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A hot-air balloon is descending at a rate of \(2.0 \mathrm{m} / \mathrm{s}\) when a passenger drops a camera. If the camera is \(45 \mathrm{m}\) above the ground when it is dropped, (a) how long does it take for the camera to reach the ground, and (b) what is its velocity just before it lands? Let upward be the positive direction for this problem.

Short Answer

Expert verified
(a) 3.24 s, (b) 29.75 m/s downward.

Step by step solution

01

Identify known values

The initial velocity of the camera is the same as the balloon's descending speed, which is \(-2.0 \mathrm{m/s}\) (since descending means downward, opposite of positive direction). The initial height \(s_0\) is \(45 \mathrm{m}\), and the acceleration due to gravity \(a\) is \(9.8 \mathrm{m/s}^2\) in the downward direction.
02

Set up motion equations

The equation for the camera's position as a function of time is \(s(t) = s_0 + v_0 t + \frac{1}{2} a t^2\), where \(s_0 = 45 \mathrm{m}\), \(v_0 = -2.0 \mathrm{m/s}\), and \(a = 9.8 \mathrm{m/s}^2\). We need to find when \(s(t) = 0\), which is when the camera hits the ground.
03

Solve for time to reach ground

Set the position equation \(s(t) = 0\):\[0 = 45 - 2t + \frac{1}{2} (9.8) t^2\] Rearrange to form a quadratic equation:\[4.9t^2 - 2t - 45 = 0\] Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4.9\), \(b = -2\), and \(c = -45\).
04

Calculate time using the quadratic formula

Plug in the values into the quadratic formula:\[t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 4.9 \times (-45)}}{2 \times 4.9}\] This simplifies to:\[t = \frac{2 \pm \sqrt{4 + 882}}{9.8}\] \[t = \frac{2 \pm \sqrt{886}}{9.8}\] Calculate the positive root:\[t \approx \frac{2 + 29.76}{9.8} \approx 3.24 \mathrm{s}\]So, it takes approximately \(3.24 \mathrm{s}\) for the camera to reach the ground.
05

Determine velocity just before impact

Use the formula for velocity: \(v = v_0 + at\).Here, \(v_0 = -2.0 \mathrm{m/s}\), \(a = 9.8 \mathrm{m/s}^2\), and \(t = 3.24 \mathrm{s}\):\[v = -2 + 9.8 \times 3.24\]\[v = -2 + 31.752\]\[v \approx 29.752 \mathrm{m/s}\]Thus, the velocity just before impact is approximately \(29.75 \mathrm{m/s}\), directed downward.

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Key Concepts

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

Kinematics
Kinematics is the study of motion without considering the forces that cause it. When analyzing motion like that of the camera falling from a hot-air balloon, we focus on displacement, velocity, and acceleration.
- **Displacement** measures how far an object has moved from its initial position. In our exercise, the initial displacement is the height from which the camera is dropped, 45 meters above the ground. - **Velocity** is the rate of change of displacement. Initial velocity is given due to the camera moving with the balloon, descending at 2 m/s. - **Acceleration** is any change in velocity over time. Here, acceleration due to gravity influences how the camera speeds up as it falls. Understanding these terms helps us formulate the equations used to predict how long the camera takes to reach the ground and its speed upon impact.
Quadratic Equation
Quadratic equations are fundamental in determining the time it takes for the camera to land. This equation arises from our kinematic equation for motion: \[ s(t) = s_0 + v_0 t + \frac{1}{2} a t^2 \]Here, inserting values converts it to:\[ 4.9t^2 - 2t - 45 = 0 \]This is a standard form quadratic equation \(ax^2 + bx + c = 0\) and can be solved using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Solving gives us two potential times, but only the positive value is realistic in our context, as time cannot be negative.
Quadratic equations frequently appear in physics for calculating time, displacement, and velocity when acceleration is constant.
Gravity
Gravity is a constant acceleration acting on objects on Earth, approximately \(9.8 \, \text{m/s}^2\) downwards. In the problem, gravity pulls the camera towards the ground, making the fall faster over time. The acceleration due to gravity is always downward, uniformly affecting projectiles like cameras, regardless of their initial velocity.When objects are freely falling, gravity is the sole force acting on them, making them accelerate. This acceleration is crucial for calculating the time of flight and the final velocity.In projectile motion, understanding gravity helps explain why the camera's speed increases as it falls and predicts the motion's overall shape and behavior.
Initial Velocity
Initial velocity matters as it affects how the object falls. In our scenario, the camera's initial velocity is negative because it is dropped from a descending balloon. This means it starts with a downward velocity of \(-2.0 \text{ m/s}\).Initial velocity is combined with gravitational acceleration to determine the camera's final speed and time taken to hit the ground.- If the camera started from rest, its initial velocity would be \(0\).- A downward initial velocity speeds up the fall due to gravity's added effect in the same direction. The initial velocity provides a starting speed for the camera, which is built upon by gravity to calculate its speed just before impact using the formula:\[ v = v_0 + at \]This showcases how initial conditions impact the future behavior of moving objects.

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