# Falling Body Problems

Consider a vertically falling body of mass ${m}$ that is being influenced only by gravity ${g{}}$ and an air resistance that is proportional to the velocity of the body. Assume that both gravity and mass remain constant and, for convenience, choose the downward direction as the positive direction.

Newton's second law of motion: The net force acting on a body is equal to the time rate of change of the momentum of the body; or, for constant mass, ${F}={m}\frac{{{d}{v}}}{{{d}{t}}}$.

For the problem at hand, there are two forces acting on the body: the force due to gravity given by the weight ${w}$ of the body, which equals ${m}{g{}}$, and the force due to air resistance given by $-{k}{v}$, where ${k}>{0}$ is a constant of proportionality. The minus sign is required because this force opposes the velocity; that is, it acts in the upward, or negative, direction. The net force ${F}$ on the body is, therefore, ${F}={m}{g{-}}{k}{v}$.
Substituting this result, we obtain ${m}{g{-}}{k}{v}={m}\frac{{{d}{v}}}{{{d}{t}}}$, or $\frac{{{d}{v}}}{{{d}{t}}}+\frac{{k}}{{m}}{v}={g{}}$, as the equation of motion for the body.
If air resistance is negligible or nonexistent, ${k}={0}$, and the equation simplifies to $\frac{{{d}{v}}}{{{d}{t}}}={g{}}$.
When ${k}>{0}$, limiting velocity ${v}_{{l}}$ is defined by ${v}_{{l}}=\frac{{{m}{g}}}{{k}}$.

The differential equation is valid only if the given conditions are satisfied. This equation is not valid, if, for example, air resistance is not proportional to velocity but to velocity squared or if the upward direction is taken to be the positive direction.

Example 1. A steel ball weighing 2 lb. is dropped from a height of 3000 ft. with no velocity. As it falls, the ball encounters an air resistance numerically equal to $\frac{{v}}{{8}}$ (in pounds), where ${v}$ denotes the velocity of the ball (in feet per second). Find the limiting velocity for the ball and the time required for the ball to hit the ground.

Locate the coordinate system with the ground now situtated at ${x}={3000}$. Here, ${w}={2}$ lb., and ${k}=\frac{{1}}{{8}}$. Assuming gravity ${g{}}$ is ${32}\frac{{f{{t}}}}{{{s}}^{{2}}}$, from the formula ${w}={m}{g{}}$, we have that ${m}=\frac{{w}}{{{g}}}=\frac{{2}}{{32}}=\frac{{1}}{{16}}$ slug. So, the differential equation is $\frac{{{d}{v}}}{{{d}{t}}}+\frac{{\frac{{1}}{{8}}}}{{\frac{{1}}{{16}}}}{v}={32}$, or $\frac{{{d}{v}}}{{{d}{t}}}+{2}{v}={32}$.

This is a first-order linear differential equation. Its solution is ${v}{\left({t}\right)}={c}{{e}}^{{-{2}{t}}}+{16}$.

We are given that initially there is no velocity; so, ${v}{\left({0}\right)}={0}$, and ${0}={c}{{e}}^{{-{2}\cdot{0}}}+{16}$, or ${c}=-{16}$.

The differential equation becomes ${v}{\left({t}\right)}=-{16}{{e}}^{{-{2}{t}}}+{16}$.

From this equation, we see that ${v}\to{16}$ when ${t}\to\infty$; so, the limiting velocity is ${16}\frac{{f{{t}}}}{{{s}}^{{2}}}$.

To find the time it takes for the ball to hit the ground $\left({x}={3000}\right)$, we need an expression for the position of the ball at any time ${t}$. Since ${v}=\frac{{{d}{x}}}{{{d}{t}}}$, we have that$\frac{{{d}{x}}}{{{d}{t}}}=-{16}{{e}}^{{-{2}{t}}}+{16}$. Integrating both sides gives ${x}={8}{{e}}^{{-{2}{t}}}+{16}{t}+{c}_{{1}}$. To find the constant, use the fact that ${x}={0}$ when ${t}={0}$: ${0}={8}{{e}}^{{-{2}\cdot{0}}}+{16}\cdot{0}+{c}_{{1}}$, or ${c}_{{1}}=-{8}$.

So, the position of the particle is given as ${x}{\left({t}\right)}={8}{{e}}^{{-{2}{t}}}+{16}{t}-{8}$.

We need to find such ${t}$ that ${x}{\left({t}\right)}={3000}$: ${3000}={8}{{e}}^{{-{2}{t}}}+{16}{t}-{8}$, or ${376}={{e}}^{{-{2}{t}}}+{2}{t}$.

This equation cannot be solved explicitly for ${t}$; we can approximate the solution by trial and error, substituting different values of ${t}$ until we locate a solution to the degree of accuracy we need. Alternatively, we note that for any large value of ${t}$, the negative exponential term will be negligible. A good approximation is obtained by setting ${2}{t}={376}$, or ${t}={188}$ seconds. For this value of ${t}$, the exponential is essentially zero.

And now, one more example.

Example 2. A body is propelled straight up with an initial velocity of ${500}\frac{{f{{t}}}}{{\sec{}}}$ in a vacuum with no air resistance. How long will it take for the body to return to the ground?

Since there is no air resistance, ${k}={0}$; so, $\frac{{{d}{v}}}{{{d}{t}}}={g{=}}{32}$, and ${v}={32}{t}+{C}$.

Since ${v}{\left({0}\right)}=-{500}$ (because we are moving up in the negative direction), we have that $-{500}={32}\cdot{0}+{C}$, or ${C}=-{500}$.

So, ${v}={32}{t}-{500}$.

It is known that $\frac{{{d}{x}}}{{{d}{t}}}={v}={32}{t}-{500}$.

Integrating this equation gives ${x}{\left({t}\right)}={16}{{t}}^{{2}}-{500}{t}+{C}_{{1}}$.

Since ${x}{\left({0}\right)}={0}$, we have that ${0}={16}\cdot{{0}}^{{2}}-{500}\cdot{0}+{C}_{{1}}$, or ${C}_{{1}}={0}$.

So, ${x}{\left({t}\right)}={16}{{t}}^{{2}}-{500}{t}$.

We need to find the time when ${x}{\left({t}\right)}={0}$.

${16}{{t}}^{{2}}-{500}{t}={0}$, or ${t}{\left({16}{t}-{500}\right)}={0}$, which gives ${t}=\frac{{500}}{{16}}={31.25}$ seconds.