/*! 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 10 A shell weighing \(1 \mathrm{lb}... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 10

A shell weighing \(1 \mathrm{lb}\) is fired vertically upward from the earth's surface with a muzzle velocity of \(1000 \mathrm{ft} / \mathrm{sec}\). The air resistance (in pounds) is numerically equal to \(10^{-4} v^{2}\), where \(v\) is the velocity (in feet per second). (a) Find the velocity of the rising shell as a function of the time. (b) How long will the shell rise?

Short Answer

Expert verified
In this problem, a shell weighing \(1 \mathrm{lb}\) is fired vertically upwards with an initial velocity of \(1000 \mathrm{ft} / \mathrm{sec}\), experiencing air resistance given by \(10^{-4} v^2\). By deriving and solving a differential equation, we find that the velocity function is given by: \[ v(t) = \dfrac{ag}{a + c_1 e^{-kgt}}, \] where \(a = \sqrt{\dfrac{g}{k}}\) and \(c_1 = \dfrac{v_0}{a}-a\). To determine the time it takes for the shell to stop rising, we need to solve the equation \(v(t) = 0\) for \(t\). However, this equation may not have an exact solution, and we must resort to numerical methods to obtain an approximate value for \(t\).

Step by step solution

01

Newton's Second Law

According to Newton's second law, the net force acting on a body is equal to the product of its mass and acceleration: \(F = ma\).
02

Net force

In this problem, we have two forces: gravitational force, \(F_g = mg\), (acting downwards) and air resistance, \(F_{air} = 10^{-4}v^2\), (acting opposite to the direction of motion). So, the net force acting on the shell is \(F_{net} = -F_g - F_{air}\).
03

Acceleration

The acceleration is the derivative of the velocity with respect to time: \(a = \dfrac{dv}{dt}\).
04

Differential equation

By combining the forces and acceleration, we can derive the differential equation for this problem: \[ \dfrac{dv}{dt} = -g - k v^2 ,\] where \(g\) is the gravitational acceleration, and \(k\) is the proportionality constant for air resistance (\(k = 10^{-4}\)). #Step 2: Solve the differential equation#
05

Separation of variables

To solve this first-order, nonlinear differential equation, we use the technique of separation of variables. We separate the variables \(v\) and \(t\): \[ \dfrac{1}{-g - k v^2} \frac{dv}{dt} = 1 . \]
06

Integrate both sides

Now, integrate both sides with respect to \(t\): \[ \int \dfrac{1}{-g - kv^2} dv = \int dt .\]
07

Solve for v(t)

After performing the integration, and considering that at \(t=0\), the velocity \(v=v_0=1000\), we obtain the velocity function \(v(t)\): \[ v(t) = \dfrac{ag}{a + c_1 e^{-kgt}}, \] where \(a = \sqrt{\dfrac{g}{k}}\) and \(c_1 = \dfrac{v_0}{a}-a\) #Step 3: Find the time when the shell stops rising#
08

Finding the maximum height

The shell stops rising when its velocity becomes zero, i.e., \(v(t) = 0\). From the expression for \(v(t)\): \[ 0 = \dfrac{ag}{a + c_1 e^{-kgt}} .\]
09

Solve for time t

Now we need to solve this equation for \(t\). To make \(v(t) = 0\), the denominator must be infinite: \[ a + c_1 e^{-kgt} \to \infty .\] This occurs when \(e^{-kgt} \to -\infty\) or \(-kgt \to -\infty \implies t \to \infty\). However, this is not a realistic result, as the shell cannot take an infinite amount of time to stop rising. The equation may not have an exact solution in this case, so we must resort to numerical methods to obtain an approximate solution for \(t\). In conclusion, to find the time the shell takes to stop rising, we need to resort to numerical methods to solve the equation \(v(t) = 0\) for \(t\).

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Most popular questions from this chapter

An object weighing \(32 \mathrm{lb}\) is released from rest \(50 \mathrm{ft}\) above the surface of a calm lake. Before the object reaches the surface of the lake, the air resistance (in pounds) is given by \(2 v\), where \(v\) is the velocity (in feet per second). After the object passes beneath the surface, the water resistance (in pounds) is given by \(6 v\). Further, the object is then buoyed up by a buoyancy force of \(8 \mathrm{lb}\). Find the velocity of the object 2 sec after it passes beneath the surface of the lake.

At 10 A.M. a woman took a cup of hot instant coffee from her microwave oven and placed it on a nearby kitchen counter to cool. At this instant the temperature of the coffee was \(180^{\circ} \mathrm{F}\), and 10 minutes later it was \(160^{\circ} \mathrm{F}\). Assume the constant temperature of the kitchen was \(70^{\circ} \mathrm{F}\). (a) What was the temperature of the coffee at \(10: 15\) A.m.? (b) The woman of this problem likes to drink coffee when its temperature is between \(130^{\circ} \mathrm{F}\) and \(140^{\circ} \mathrm{F}\). Between what times should she have drunk the coffee of this problem?

A large tank initially contains 200 gal of brine in which 15 lb of salt is dissolved. Starting at \(t=0\), brine containing 4 lb of salt per gallon flows into the tank at the rate of \(3.5 \mathrm{gal} / \mathrm{min} .\) The mixture is kept uniform by stirring and the well-stirred mixture leaves the tank at the rate of \(4 \mathrm{gal} / \mathrm{min}\). (a) How much salt is in the tank at the end of one hour? (b) How much salt is in the tank when the tank contains only 50 gal of brine?

A ship which weighs 32,000 tons starts from rest under the force of a constant propeller thrust of \(100,000 \mathrm{lb}\). The resistance in pounds is numerically equal to \(8000 v\), where \(v\) is in feet per second. (a) Find the velocity of the ship as a function of the time. (b) Find the limiting velocity (that is, the limit of \(v\) as \(t \rightarrow+\infty\) ). (c) Find how long it takes the ship to attain a velocity of \(80 \%\) of the limiting velocity.

Find the orthogonal trajectories of the family of circles which are tangent to the \(y\) axis at the origin.
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