17. Applications of Differential Equations

e. Falling near the Earth

2. Falling with Air Resistance

Air resistance is a drag force which pushes back on the body, opposite to the velocity, slowing it down. Further, the drag increases as the speed increases. So the drag force is usually taken to be \[ F=-kv \] where \(k\) is a positive constant called the drag coefficient or the coefficient of viscosity. There is a minus sign because if the body is moving down so that the velocity \(v\) is negative then the force is positive and pushes up. The drag coefficient depends on the size and shape of the body as well as on the density of the air (or other fluid).

The total force is now the sum of the gravitational force and the drag force. So Newton's Second Law now says \(ma=-mg-kv\). Consequently, the position of the body \(y(t)\) may now be found by solving the differential equation: \[ m\dfrac{d^2y}{dt^2}=-mg-k\dfrac{dy}{dt} \] This equation is most easily solved by first finding the velocity \(v(t)\) by solving the velocity equation: \[ m\dfrac{dv}{dt}=-mg-kv \] and then finding the position \(y(t)\) by solving the position equation: \[ \dfrac{dy}{dt}=v \] The velocity equation is both separable and linear, and so may be solved by either separating variables or by multiplying by an integrating factor. The position equation may be solved by a simple integration.

Further, we will see that, when there is air resistance, the velocity approaches a terminal velocity: \[\begin{aligned} v_\text{terminal} &=\lim_{t\rightarrow\infty}v(t) \end{aligned}\]

A soccer ball whose mass is \(0.5\) kg is dropped (\(v(0)=0\)) from the top of the Leaning Tower of Pisa which is \(55\) m tall. If the drag coefficient is \(k=0.8\,\dfrac{\text{kg}}{\text{m}}\), how long does it take the ball to reach the ground? What is its velocity when it reaches the ground? Graph its position and velocity as a function of time. What is the terminal velocity and approximately when does the ball reach terminal velocity?

The velocity equation is \[ .5\dfrac{dv}{dt}=-.5(9.8)-.8v \] We multiply through by \(2\) and write this in standard linear form: \[ \dfrac{dv}{dt}+1.6v=-9.8 \] The integrating factor is \[ I=e^{\textstyle \int 1.6\,dt}=e^{1.6t} \] We multiply the standard form of the differential equation by the integrating factor, recognize the left as the derivative of a product and integrate: \[\begin{aligned} e^{1.6t}\dfrac{dv}{dt}+e^{1.6t}1.6v&=-9.8e^{1.6t} \\ \dfrac{d}{dt}\left[e^{1.6t}v\right]&=-9.8e^{1.6t} \end{aligned}\] \[\begin{aligned} e^{1.6t}v&=-\int 9.8e^{1.6t}\,dt=-\dfrac{9.8}{1.6}e^{1.6t}+C \\ &=-6.125e^{1.6t}+C \end{aligned}\] We solve for \(v\): \[ v=-6.125+Ce^{-1.6t} \] Next, we use the initial condition \(v=0\) when \(t=0\): \[ 0=-6.125+C \qquad \Longrightarrow \qquad C=6.125 \] So the velocity is \(v=-6.125+6.125e^{-1.6t}\) or: \[ v=-6.125\left(1-e^{-1.6t}\right) \] Notice that the velocity is negative for \(t \gt 0\). So the speed (the absolute value of the velocity) is: \[ \text{speed}\,=|v|=6.125\left(1-e^{-1.6t}\right) \] Finally, we get the position by integrating the velocity: \[\begin{aligned} y&=\int v\,dt=\int \left(6.125e^{-1.6t}-6.125\right)\,dt \\ &=-\dfrac{6.125}{1.6}e^{-1.6t}-6.125t+K \\ &=-3.828e^{-1.6t}-6.125t+K \end{aligned}\] and using the initial condition \(y=55\) when \(t=0\): \[ 55=-3.828+K \qquad \Longrightarrow \qquad K=58.828 \] So the position is \(y=-3.828e^{-1.6t}-6.125t+58.828\) or: \[ y=55+3.828\left(1-e^{-1.6t}\right)-6.125t \]

The plots show that the ball reaches the terminal velocity at about \(t=3\) sec where the speed graph becomes horizontal and the position graph becomes a straight line.