17. Applications of Differential Equations

e. Falling near the Earth

On this and the next page, we study the motion of a body falling vertically in the gravitational field near the surface of the Earth. For this purpose, we will take the \(y\)-axis as vertical with the positive direction pointing up. We take the position as \(y(t)\). Then the velocity is \(v(t)=\dfrac{dy}{dt}\) and the acceleration is \(a(t)=\dfrac{dv}{dt}=\dfrac{d^2y}{dt^2}\). When a body of mass \(m\) falls, it feels a constant gravitational force \(F=-mg\) where \[ g=9.8\,\dfrac{\text{m}}{\text{sec}^2} =980\,\dfrac{\text{cm}}{\text{sec}^2} =32\,\dfrac{\text{ft}}{\text{sec}^2} \] is called the acceleration of gravity. There is a minus sign because the force pulls down. Newton's Second Law says \(F=ma\), where \(F\) is the sum of all forces acting on the body. It is frequently useful to reverse this, writing \(ma=F\) or \[ m\dfrac{d^2y}{dt^2}=F \] and use this second order differential equation to solve for the position, \(y(t)\). The solution method is to split this into two first order equations: \[ \dfrac{dy}{dt}=v(t) \qquad \text{and} \qquad \dfrac{dv}{dt}=\dfrac{1}{m}F \] Given the total force, \(F\), we integrate the second equation, called the velocity equation, to find the velocity, \(v(t)\), and then integrate the first equation, called the position equation, to find the position, \(y(t)\).

1. Free Fall

We say a body is in free fall when the only force is gravity, \(F=-mg\). Consequently, the position of the body \(y(t)\) may be found by solving the differential equation: \[ m\dfrac{d^2y}{dt^2}=-mg \] After cancelling the \(m\), the solution is easily found by integrating twice: \[\begin{aligned} \dfrac{dy}{dt}&=-gt+v_o \\ y&=-\dfrac{1}{2}gt^2+v_ot+y_o \end{aligned}\] where \(y_o\) and \(v_o\) are the constants of integration which may be interpreted as the initial position and velocity, since: \[ y(0)=y_o \qquad \text{and} \qquad v(0)=\dfrac{dy}{dt}(0)=v_o \]

Find the amount of time it takes a golf ball to fall to the ground if it is dropped from \(4\) m starting from rest, if the only force is gravity. What is its velocity when it hits the ground?

"Dropped from \(4\) m starting from rest" means the initial conditions are \(y(0)=4\) and \(v(0)=0\).

The golf ball hits the ground at: \(t_g=\sqrt{\dfrac{4}{490}}\,\text{sec}\,\approx.09\,\text{sec} \)
and its velocity is: \(v(t_g)=-980\sqrt{\dfrac{4}{490}}\,\dfrac{\text{m}}{\text{sec}} \approx88.5\,\dfrac{\text{m}}{\text{sec}} \)

The initial conditions are \(y(0)=y_o=4\) and \(v(0)=v_o=0\). We integrate the acceleration equation: \[ \dfrac{dv}{dt}=-g=-9.8 \] to get the velocity \[ v(t)=-9.8t+v_o=-9.8t \] We integrate the position equation: \[ \dfrac{dy}{dt}=v(t)=-9.8t \] to get the position \[ y(t)=-4.9t^2+y_o=-4.9t^2+4 \] The ball hits the ground at a time \(t_g\) when \(y(t_g)=0\), or: \[ -4.9t_g^2+4=0 \qquad \Longrightarrow \qquad t_g=\sqrt{\dfrac{4}{4.9}}\approx.9\,\text{sec} \] When it hits the ground, its velocity is: \[ v(t_g)=-9.8t_g=-9.8\sqrt{\dfrac{4}{4.9}}\approx8.85\,\dfrac{\text{m}}{\text{sec}} \]

The above computations assume that gravity is the only force acting on the body. In reality, there are other forces, most typically, the drag force of air resistance discussed on the next page.

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Supported in part by NSF Grant #1123255