Question

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

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

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: \(F = ma\).

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 \(F_{net} = -F_g - F_{air}\).

Acceleration

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

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#

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 \(t\): \[ \dfrac{1}{-g - k v^2} \frac{dv}{dt} = 1 . \]

Integrate both sides

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

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#

Finding the maximum height

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

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 \(t\).