/*! 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 11 A projectile of mass \(100 \math... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 11

A projectile of mass \(100 \mathrm{~kg}\) is launched vertically from ground level with an initial velocity of \(100 \mathrm{~m} / \mathrm{s}\). (a) If the projectile is subjected to air resistance equivalent to \(1 / 10\) times the instantaneous velocity, determine the velocity of and the height of the projectile at any time \(t\). (b) What is the maximum height attained by the projectile?

Short Answer

Expert verified
The velocity is \(v(t) = 100 e^{-t/10} - 100\) and the height is \(h(t) = -1000e^{-t/10} + 100t + 1000\). The maximum height is the peak of the height function obtained.

Step by step solution

01

- Understanding the Problem

Identify the given quantities and what is being asked. The mass of the projectile is given as 100 kg, the initial velocity is 100 m/s, and the air resistance is 1/10 of the instantaneous velocity. Determine the velocity and height of the projectile as functions of time, and find the maximum height.
02

- Writing the Differential Equation

The force of air resistance is proportional to the velocity, which can be expressed as: \[ F_{air} = -\frac{1}{10}v \] The net force on the projectile is then: \[ F_{net} = m \frac{dv}{dt} = -mg - \frac{v}{10} \]where \(g\) is the acceleration due to gravity (9.8 m/s虏).
03

- Setting Up the Differential Equation

Using Newton's second law: \[ m \frac{dv}{dt} = -mg - \frac{v}{10} \]Simplify the equation: \[ \frac{dv}{dt} + \frac{v}{10} = -g \]
04

- Solution to the Differential Equation for Velocity

Solve the first-order linear differential equation. The integrating factor is \[ \text{IF} = e^{\frac{t}{10}} \]Multiplying both sides of the equation by the integrating factor: \[ e^{\frac{t}{10}} \frac{dv}{dt} + \frac{v}{10}e^{\frac{t}{10}} = -ge^{\frac{t}{10}} \]Integrate both sides: \[ v e^{\frac{t}{10}} = \frac{-g}{10} e^{\frac{t}{10}} + C \]Solve for v: \[ v(t) = (C - \frac{100g}{10}) e^{-\frac{t}{10}} \]Using the initial condition \(v(0) = 100\): \[ 100 = (C - 100) \]So, \[ C = 200 \]Thus, \[ v(t) = (200 - 100) e^{-\frac{t}{10}} - 100 = 100 e^{-\frac{t}{10}} - 100 \]
05

- Find the Height/Position Function

Integrate the velocity function to find the position function: \[ h(t) = \frac{d}{dt} \bigg( -1000e^{-\frac{t}{10}} + 100t + h_0 \bigg) \]Since the projectile starts at ground level, \(h(0) = 0\): \[ h(0) = -1000 + h_0 = 0 \]Thus, \(h_0 = 1000\): \[ h(t) = -1000e^{-\frac{t}{10}} + 100t + 1000 \]
06

- Maximum Height Calculation

To find the maximum height, set \(v(t) = 0\): \[ 100e^{-\frac{t}{10}} - 100 = 0 \]Hence, \[ e^{-\frac{t}{10}} = 1 \]Thus, \(t\) is infinite, but because of acceleration, this only considers the more theoretically grounded maximum height \(t\) occurs very close to initial motion h is: \[ h_{max} = -1000e^{-\frac{t}{10}} + 100t + 1000 \]

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Key Concepts

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

Air Resistance
Air resistance plays a crucial role in projectile motion as it acts opposite to the direction of the projectile's velocity. It can significantly affect the trajectory and maximum height of the projectile. The drag force can be modeled as proportional to the velocity of the projectile. In this exercise, the air resistance is given as \[ F_{air} = -\frac{1}{10}v \]. This means that for every speed the projectile achieves, there is a counteracting force reducing it by 1/10 of the velocity.
First-Order Linear Differential Equations
Solving for the velocity of the projectile involves first-order linear differential equations. These types of equations are expressed in the form \[ \frac{dv}{dt} + P(t)v = Q(t) \]. In our exercise, the differential equation for the velocity under air resistance is derived as \[ \frac{dv}{dt} + \frac{v}{10} = -g \]. The goal is to solve this equation for v, the velocity, as a function of time.
Maximum Height Calculation
To determine the maximum height, we set the velocity to zero, since the highest point in the projectile's motion is reached when its velocity changes direction (i.e., becomes zero). From the velocity solution \[ v(t) = 100e^{-\frac{t}{10}} - 100 \], setting \[v(t) = 0 \] helps us find the time at which the maximum height is reached. Substituting this time back into the height function \[ h(t) \] provides the maximum height attained by the projectile.
Newton's Second Law
Newton's second law of motion states that the force acting on an object is equal to the mass of the object multiplied by its acceleration: \[ F = ma \]. In this projectile motion exercise, the forces include gravity and air resistance. Thus, the net force \[ F_{net} = m \frac{dv}{dt} \] accounts for both gravity (\[ -mg \]) and air resistance (\

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Most popular questions from this chapter

Consider a radioactive substance with halflife 10 days. If there are initially \(5000 \mathrm{~g}\) of the substance, how much remains after 365 days?

Suppose that an object of mass \(10 \mathrm{~kg}\) is thrown vertically into the air with an initial velocity of \(v_{0} \mathrm{~m} / \mathrm{s}\). If the limiting velocity is \(-19.6 \mathrm{~m} / \mathrm{s}\), what can be said about \(c>0\) (positive constant), in the force due to air resistance \(F_{R}=c v\) acting on the object?

(Harvesting) If we wish to model a population of size \(P(t)\) at time \(t\) and consider a constant harvest rate \(h\) (like hunting, fishing, or disease), then we might modify the logistic equation and use the equation \(P^{\prime}=\) \(r P-a P^{2}-h\) to model the population under consideration. Assume that \(h \geq r^{2} /(4 a)\). (a) Show that if \(h \geq r^{2} /(4 a)\), a general solution of \(P^{\prime}=r P-a P^{2}-h\) is $$ \begin{aligned} P(t) &=\frac{1}{2 a}\left[r+\sqrt{4 a h-r^{2}}\right.\\\ &\left.\times \tan \left(\frac{1}{2 a}(C-a t) \sqrt{4 a h-r^{2}}\right)\right] . \end{aligned} $$ (b) Suppose that for a certain species it is found that \(r=0.03, a=0.0001, h=2.26\), and \(C=-1501.85\). At what time will the species become extinct? (c) If \(r=0.03, a=0.0001\), and \(P(0)=5.3\), graph \(P(t)\) if \(h=0,0.5,1.0,1.5,2.0,2.25\) and 2.5. (d) What is the maximum allowable harvest rate to assure that the species survives? (e) Generalize your result from (d). For arbitrary \(a\) and \(r\), what is the maximum allowable harvest rate that ensures survival of the species?

In a chemical reaction, chemical \(\mathrm{A}\) is converted to chemical \(B\) at a rate proportional to the amount of chemical A present. If half of chemical A remains after five hours, when does \(1 / 6\) of the initial amount of chemical A remain? How much of the initial amount remains after \(15 \mathrm{~h}\) ?

A parachutist weighing \(60 \mathrm{~kg}\) falls from a plane (that is, \(v_{0}=0\) ) and is subjected to air resistance equivalent to \(F_{R}=10 v\). After one minute, the parachute is opened so that the parachutist is subjected to air resistance equivalent to \(F_{R}=100 \mathrm{v}\). (a) What is the parachutist's velocity when the parachute is opened? (b) What is the parachutist's velocity \(v(t)\) after the parachute is opened? (c) What is the parachutist's limiting velocity? How does this compare to the limiting velocity if the parachute does not open?
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