/*! 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 133 A skydiver is at an altitude of ... [FREE SOLUTION] | 91Ó°ÊÓ

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A skydiver is at an altitude of 1520 m. After 10.0 seconds of free fall, he opens his parachute and finds that the air resistance, \(F_{D},\) is given by the formula \(F_{\mathrm{D}}=-b v, \quad\) where \(b\) is a constant and \(v\) is the velocity. If \(b=0.750,\) and the mass of the skydiver is \(82.0 \mathrm{kg}\) first set up differential equations for the velocity and the position, and then find: (a) the speed of the skydiver when the parachute opens, (b) the distance fallen before the parachute opens, (c) the terminal velocity after the parachute opens (find the limiting velocity), and (d) the time the skydiver is in the air after the parachute opens.

Short Answer

Expert verified
In summary, for the given skydiver and parachute problem, we obtain the following results: a) The speed when the parachute opens is \(98.1 \, m/s\). b) The distance fallen before the parachute opens is \(490.5 \, m\). c) The terminal velocity after the parachute opens is \(1073.2 \, m/s\). d) The time in the air after the parachute opens is \(0.959 \, s\).

Step by step solution

01

Set up differential equations for velocity and position

We are given that the air resistance force, \(F_D\) is given by the formula \(F_D = -bv\), where \(b\) is a constant and \(v\) is the velocity. Using Newton's second law of motion, the net force acting on the skydiver is given by: \(F_{net} = F_D + F_g = ma\), where \(F_g = mg\) is the gravitational force acting on the skydiver, and \(a\) is skydiver's acceleration. Since \(F_D = -bv, F_{net} = -bv + mg\). Now we substitute the acceleration \(a\) with the derivative of velocity with respect to time, \(a=\frac{dv}{dt}\) in the equation above: \(\frac{dv}{dt} = \frac{-bv + mg}{m}\) For position, we know that the derivative of position with respect to time is velocity, or \(\frac{ds}{dt} = v\). Now, we have two differential equations: 1. \(\frac{dv}{dt} = \frac{-bv + mg}{m}\) 2. \(\frac{ds}{dt} = v\)
02

Find the speed when the parachute opens

We are given that the skydiver free falls for 10 seconds. In the free fall phase, the air resistance force \(F_D = 0\), so the net force is equal to gravitational force, \(F_{net} = mg\). In this case, the differential equation for velocity becomes: \(\frac{dv}{dt} = g\) Integrating the differential equation, we get \(v = gt + C_1\). At the beginning, \(t = 0\) and \(v = 0\). Using these initial conditions, we find \(C_1 = 0\), and the equation becomes \(v = gt\). Now we can find the speed when the parachute opens by substituting the time \(t = 10s\) into the equation: \(v = g(10) = 9.81(10) = 98.1 \, m/s\)
03

Find the distance fallen before the parachute opens

Using the differential equation for position, \(\frac{ds}{dt} = v\), and the velocity equation \(v = gt\), we get: \(\frac{ds}{dt} = gt\) Integrating the differential equation with respect to time, we get \(s = \frac{1}{2}gt^2 + C_2\). At the beginning, \(t = 0\) and \(s = 0\). Using these initial conditions, we find \(C_2 = 0\), and the equation becomes \(s = \frac{1}{2}gt^2\). Now we can find the distance fallen before the parachute opens by substituting the time \(t = 10s\) into the equation: \(s = \frac{1}{2}(9.81)(10)^2 = 490.5 \, m\)
04

Find the terminal velocity after the parachute opens

To find the terminal velocity, we need to find the limiting velocity when the air resistance force equals the gravitational force, or when \(F_D = -F_g\). As \(F_D = -bv\), we have: \(-bv = -mg\) Now we can find terminal velocity, \(v_t\), by rearranging the equation: \(v_t = \frac{mg}{b}\) Substituting given values: \(v_t = \frac{(82.0)(9.81)}{0.750} = 1073.2\, m/s\)
05

Find the time in the air after the parachute opens

To find the time in the air after the parachute opens, we can consider the altitude when the parachute opens and when the skydiver reaches the ground. Since the skydiver started at an altitude of 1520 m and fall 490.5 m before the parachute opens, the remaining altitude after the parachute opens is \(1520 - 490.5 = 1029.5\, m\). Since the terminal velocity, \(v_t = 1073.2\, m/s\), is constant and the skydiver travels 1029.5 m at that velocity, we can find the time, \(t_{air}\), as follows: \(t_{air} = \frac{1029.5}{1073.2} = 0.959\, s\) To recap, we find: a) The speed when the parachute opens: 98.1 m/s. b) The distance fallen before the parachute opens: 490.5 m. c) The terminal velocity after the parachute opens: 1073.2 m/s. d) The time in the air after the parachute opens: 0.959 s.

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Key Concepts

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

Newton's Second Law
Newton's Second Law of Motion is fundamental to understanding the skydiver's motion. It can be stated as: the force acting on an object is equal to the mass of that object multiplied by its acceleration. Expressed mathematically, this law is written as \( F = ma \), where \( F \) is the net force applied, \( m \) is the mass, and \( a \) is the acceleration. In the context of a skydiver, this law helps us calculate how the forces of gravity and air resistance work together and influence the skydiver's velocity and acceleration. - **Gravitational Force:** The force that pulls the skydiver towards Earth, represented by \( F_g = mg \), where \( g \approx 9.81 \ m/s^2 \). - **Air Resistance (Drag):** In the problem, this is represented by \( F_D = -bv \), acting in the opposite direction of motion due to air pushing against the skydiver. The equation \( F_{net} = ma \) describes the balance of these forces, allowing us to solve for acceleration and, consequently, for the velocity of the skydiver as they fall.
Terminal Velocity
Terminal velocity occurs when a falling object reaches a point where the force of gravity is balanced by the force of air resistance, causing it to stop accelerating and maintain a constant speed.For the skydiver, terminal velocity is reached when the drag force \( F_D = -bv \) equals the gravitational force \( F_g = mg \). When these forces exactly oppose each other, we have:- \( mg = bv \)- Rearranging, the terminal velocity \(v_t\) can be calculated as \( v_t = \frac{mg}{b} \).This final speed represents the fastest velocity the skydiver will reach during the fall. It's important because knowing the terminal velocity allows us to predict the descent rate once the parachute deploys or when sufficient height is lost without acceleration due to ongoing drag.
Free Fall
Free fall describes an object falling solely under the influence of gravity, excluding significant effects from air resistance.When the skydiver initially jumps from the plane, they experience free fall since gravity is the primary force acting on them. During this phase:- **Gravity Only:** The net force equals the gravitational force, leading to the differential equation: \( \frac{dv}{dt} = g \).- **Velocity Builds:** The velocity increases steadily due to constant acceleration from gravity, with the initial condition \(v = gt\).The skydiver continues in free fall until reaching significant air resistance, like when the parachute is deployed. Computing free fall characteristics helps us determine the speed and distance covered during this part of the descent.
Air Resistance
Air resistance, or drag, is the force exerted by air against the direction of movement for a falling object. It plays a crucial role in determining the speed and acceleration of the skydiver after jumping.Air resistance depends on factors such as:- **Velocity of the Object:** As velocity increases, the force of air resistance increases. It is often modeled with the formula \( F_D = -bv \), where \( b \) is a drag coefficient and \( v \) is velocity.- **Surface Area and Shape:** Greater surface area generally increases air resistance. For skydivers, body positioning and parachute deployment significantly affect drag.In the skydiver's descent, air resistance starts insignificant but grows as velocity increases, ultimately balancing the gravitational force and leading the skydiver to reach terminal velocity. Understanding how air resistance operates is vital for solving equations of motion and accurately determining final speeds.

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