1.2: Newton Mechanics - Free Fall

Dimensions are useful not just to debunk incorrect arguments but also to generate correct ones. To do so, the quantities in a problem need to have dimensions. As a contrary example showing what not to do, here is how many calculus textbooks introduce a classic problem in motion:

A ball initially at rest falls from a height of h feet and hits the ground at a speed of v feet per second. Find v assuming a gravitational acceleration of g feet per second squared and neglecting air resistance.

The units such as feet or feet per second are highlighted in boldface because their inclusion is so frequent as to otherwise escape notice, and their inclusion creates a significant problem. Because the height is h feet, the variable h does not contain the units of height: h is therefore dimensionless. (For h to have dimensions, the problem would instead state simply that the ball falls from a height h; then the dimension of length would belong to h.) A similar explicit specification of units means that the variables g and v are also dimensionless. Because g, h, and v are dimensionless, any comparison of v with quantities derived from g and h is a comparison between dimensionless quantities. It is therefore always dimensionally valid, so dimensional analysis cannot help us guess the impact speed.

Giving up the valuable tool of dimensions is like fighting with one hand tied behind our back. Thereby constrained, we must instead solve the following differential equation with initial conditions:

\[\frac{d^{2}y}{dt^{2}} = -g, \text{ with } y(0) = h \text{ and } dy/dt = 0 \text{ at } t = 0, \label{1.1} \]

where y(t) is the ball’s height, dy/dt is the ball’s velocity, and g is the gravitational acceleration.

Problem 1.3 Calculus solution

Use calculus to show that the free-fall differential equation \(d^{2}y/dt^{2}\) = −g with initial conditions y(0) = h and dy/dt = 0 at t = 0 has the following solution:

\[\frac{dy}{dt} = -gt \text{ and } y = -\frac{1}{2}gt^{2} + h. \label{1.2} \]

Question

Using the solutions for the ball’s position and velocity in Problem 1.3, what is the impact speed?

When y(t) = 0, the ball meets the ground. Thus the impact time t is \(\sqrt{2h/g}\). The impact velocity is −gt\(_{0}\) or − \(\sqrt{2gh}\). Therefore the impact speed (the unsigned velocity) is \(\sqrt{2gh}\).

This analysis invites several algebra mistakes: forgetting to take a square root when solving for \(t_{0}\), or dividing rather than multiplying by g when finding the impact velocity. Practice in other words, making and correcting many mistakes reduces their prevalence in simple problems, but complex problems with many steps remain minefields. We would like less error-prone methods.

One robust alternative is the method of dimensional analysis. But this tool requires that at least one quantity among v, g, and h have dimensions. Otherwise, every candidate impact speed, no matter how absurd, equates dimensionless quantities and therefore has valid dimensions.

Therefore, let’s restate the free-fall problem so that the quantities retain their dimensions:

• A ball initially at rest falls from a height h and hits the ground at speed v. Find v assuming a gravitational acceleration g and neglecting air resistance.

The restatement is, first, shorter and crisper than the original phrasing:

• A ball initially at rest falls from a height of h feet and hits the ground at a speed of v feet per second. Find v assuming a gravitational acceleration of g feet per second squared and neglecting air resistance.

Second, the restatement is more general. It makes no assumption about the system of units, so it is useful even if meters, cubits, or furlongs are the unit of length. Most importantly, the restatement gives dimensions to h, g, and v. Their dimensions will almost uniquely determine the impact speed—without our needing to solve a differential equation.

The dimensions of height h are simply length or, for short, L. The dimensions of gravitational acceleration g are length per time squared or \(LT^{−2}\), where T represents the dimension of time. A speed has dimensions of \(LT^{-1}\), so v is a function of g and h with dimensions of \(LT^{-1}\).

Problem 1.4 Dimensions of familiar quantities

In terms of the basic dimensions length L, mass M, and time T, what are the dimensions of energy, power, and torque?

Question

What combination of g and h has dimensions of speed?

The combination \(\sqrt{gh}\) has dimensions of speed.

\((\underbrace{\mathrm{LT}^{-2}}_{\mathrm{g}} \times \underbrace{\mathrm{L}}_{\mathrm{h}})^{1 / 2}=\sqrt{\mathrm{L}^{2} \mathrm{~T}^{-2}}=\underbrace{\mathrm{LT}^{-1}}_{\text {speed }} .\) \[\label{1.3} \]

Question

Is \(\sqrt{gh}\) the only combination of g and h with dimensions of speed?

In order to decide whether \(\sqrt{gh}\) is the only possibility, use constraint propagation [43]. The strongest constraint is that the combination of g and h, being a speed, should have dimensions of inverse time (\(T^{−1}\)). Because h contains no dimensions of time, it cannot help construct \(T^{-1}\).

Because g contains \(T^{-2}\), the \(T^{-1}\) must come from \(\sqrt{g}\). The second constraint is that the combination contain \(L^{1}\). The \(\sqrt{g}\) already contributes \(L^{1/2}\), so the missing \(L^{1/2}\) must come from \(\sqrt{h}\). The two constraints thereby determine uniquely how g and h appear in the impact speed v.

The exact expression for v is, however, not unique. It could be (\sqrt{gh}\), (\sqrt{2gh}\), or, in general, (\sqrt{gh}\) × dimensionless constant. The idiom of multiplication by a dimensionless constant occurs frequently and deserves a compact notation akin to the equals sign:

\[v∼ \sqrt{gh} \label{1.4} \]

Including this ∼ notation, we have several species of equality:

∝ equality except perhaps for a factor with dimensions,

∼ equality except perhaps for a factor without dimensions,

≈ equality except perhaps for a factor close to 1.

The exact impact speed is \(\sqrt{2gh}\), so the dimensions result \(\sqrt{gh}\) contains the entire functional dependence! It lacks only the dimensionless factor \(\sqrt{2}\), and these factors are often unimportant. In this example, the height might vary from a few centimeters (a flea hopping) to a few meters (a cat jumping from a ledge). The factor-of-100 variation in height contributes a factor-of-10 variation in impact speed. Similarly, the gravitational acceleration might vary from 0.27 m\(s^{−2}\)(on the asteroid Ceres) to 25 m\(s^{−2}\)(on Jupiter). The factor-of-100 variation in g contributes another factor-of-10 variation in impact speed. Much variation in the impact speed, therefore, comes not from the dimensionless factor \(\sqrt{2}\) but rather from the symbolic factors which are computed exactly by dimensional analysis. Furthermore, not calculating the exact answer can be an advantage. Exact answers have all factors and terms, permitting less important information, such as the dimensionless factor such as \(\sqrt{gh}\). As William James advised, “The art of being wise is the art of knowing what to overlook” [19, Chapter 22].

Problem 1.5 Vertical throw

You throw a ball directly upward with speed v0. Use dimensional analysis to estimate how long the ball takes to return to your hand (neglecting air resistance). Then find the exact time by solving the free-fall differential equation. What dimensionless factor was missing from the dimensional-analysis result?