Falling Body with Air Resistance

Assume that a body of mass m is dropped from a great height above the surface of the earth. Our task is to answer the following

Important Question: How far does the object fall in the first 100 seconds?

We assume that there are two forces affecting the vertical descent of the object: gravity and air resistance. Here's what we are going to do:

Gravity

Gravity makes the object fall. It was Galileo who offered the hypothesis that the acceleration due to gravity near the Earth's surface is essentially constant. Newton's law, force = mass x acceleration, then implies that

Fgrav = mg

where m is the mass of the object kilograms and g is the gravitational acceleration near the earth's surface, which is about 10 m/sec2. The basic unit of force in the metric system is the newton, which is one kg*m/sec2. Both m and g are taken to be positive.

Air Resistance

Let s(t) be the downward speed of the object t seconds after it is dropped, measured in m/sec. So we are assuming that s(0) = 0. One hypothesis governing air resistance is the following:

"Linear" Air Resistance Hypothesis


The force due to air resistance is proportional to the speed, and is applied in the direction opposite to motion.

Look at it this way, as the object moves through the air, it collides with air molecules, displacing them as it falls. The faster the object moves, the more collisions and so the greater the overall force due to air resistance. An alternative hypothesis is that the force due to air resistance is proportional to the square of the speed...this might be more reasonable, for example, at very high speeds. But we will stick with the linear assumption.

This linear hypothesis means that there is a positive constant k such that

Fair = -ks(t)

for all t. The positive constant k is the constant of proportionality and its units are kg/sec; the numerical value of k depends upon the shape of the object being dropped and the density of the atmosphere.

Total Force, Acceleration, and Speed

We now see that the total force acting on the falling object at any time t is given by

Ftot = Fgrav + Fair = mg - ks(t)

Now comes the first real conclusion involving calculus. Remember that acceleration is the rate of change of speed. (Actually it's better to think of acceleration as the rate of change of velocity, but we can identify the two concepts in this example since the object only moves in one direction.) Thus, if a(t) is the acceleration on the object at time t, then a(t) = s'(t). Now we use Newton's Law to conclude that Ftot = ma(t), or more precisely, we obtain the following relationship between speed s(t) and acceleration s'(t).

Differential Equation
Relating Speed and Accleration
mg - ks(t) = ms'(t)

Solving the Differential Equation

Our differential equation can be rewritten as

s'(t) = (k/m)[(mg/k) - s(t)]

We now introduce the auxilliary function p(t) = (mg/k) - s(t), so that p'(t) = -s'(t). Rewriting the equation in terms of p, we have -p'(t) = (k/m)p(t) or

p'(t) = (-k/m)p(t)

Now we see that the auxilliary function p is proportional to its derivative. An important fact from Differential Calculus is that any function that is proportional to its own derivative must be an exponential function. More explicitly, there is a constant C such that

[Wow!!! How'd that happen?!?!

p(t) = Ce(-k/m)t.

Rewriting this in terms of the speed function s = (mg/k) - p, we find that

s(t) = (mg/k) - Ce(-k/m)t.

We can use the initial condition s(0) = 0 to determine the constant C:

0 = s(0) = (mg/k) - Ce0 = (mg/k) - C so C = mg/k.

And here we have a solution to our differential equation, which is nothing less than an explicit function that describes the speed s(t) of the falling object t seconds after the drop:

An Explicit Formula
for Speed at Time t
s(t) = (mg/k)(1 - e(-k/m)t)

A Neat Thing to Notice: Terminal Velocity

Observe that under our assumptions (the forces acting are gravity and "linear" air resistance) we discover the phenomenon of terminal velocity. This comes about by asking for the limit of speed at time goes to infinity. As t goes to infinitey, the term e(-k/m)t goes to zero since k and m are positive. So we find

terminal velocity = limt-->infty s(t) = mg/k

The phenomenon of terminal velocity is thus discovered as a consequence of our force assumptions. (!!!)

A Sample Graph of Speed vs Time

Using the specific values g = 10 m/sec2, m = 1 kg, and k = 0.1 kg/sec, the formula for the speed function and its graph look like this:

Speed vs Time with
g=10m/sec, m = 1kg, k = 0.1kg/sec
s(t) = 100(1 - e-0.1t) m/sec

The terminal velocity phenomenon shows up as a horizontal asymptote in the graph. In this case the terminal velocity is mg/k = 100 m/sec.

Calculating the Area Under the Speed Graph

Remember our Question: How far does the object fall during the first 100 seconds after being dropped? Now when an object travels at constant speed for a certain period of time, then the distance traveled is the product of the speed and the elapsed time.

distance = (constant speed) x (elapsed time)

Our problem, of course, is that a falling body under the influence of gravity and air resistance does not fall at constant speed; just note that the speed graph above is not a horizontal line. Our job now is to see that the distance fallen by the object is equal to the area under the speed graph from t=0 to t=100. And we must develop a means to calculate, or at least approximate this area.

Here is the basic strategy, and this is really the jumping off point for Integral Calculus. First, let us subdivide the time interval [0,100] into ten smaller subintervals of equal length. So the endpoints of the subintervals are

Time Subintervals
SubintervalLeft EndpointRight Endpoint
First010
Second1020
Third2030
etc....
Ninth8090
Tenth90100

While there is a large variation of speed over the entire 100 second interval (from 0m/sec to nearly 100m/sec), the variation of speed over each of the ten subintervals is considerably less. Looking at the speed graph, we see that this reduction in speed variation is particularly true between 70 and 100 seconds after the drop. The conclusion is this:

Subinterval Approximation Idea
We can approximate the distance traveled during each time subinterval with reasonable accuracy by calculating the distance that would have been traveled had the speed been constant during that interval.

To implement this idea specifically, we can estimate how far the object dropped during a given 10 second interval by calculating how far an object would travel if its speed during that interval were equal to the speed attained at the end of the time subinterval. Since the speed of the falling object is increasing, this process is guaranteed to produce an overestimate. For example, at the beginning of the fourth time subinterval, that is when t = 30, the speed is s(30) = 100(1-e-3) or about 95.0m/sec. At the end of the fourth time subinterval, the speed is s(40) = 98.2m/sec. If we then use this latter speed, we estimate that the distance traveled during the fourth time interval is about

(approx constant speed) x (elapsed time) = (98.2m/sec)(10 sec) = 982 m.

And this is guaranteed to be an overestimate. We could use the beginning speed (or left endpoint speed) to obtain an underestimate of 950m traveled. In any event, the actual distance is somewhere between these numbers.

We can make this systematic by agreeing for now to always choose the right endpoint of each time interval, and then add up the estimated subinterval distances to obtain an (over)estimate of the distance traveled during the entire 100 second span. All ending speeds below are calculated from the formula for s(t) and are rounded to the nearest 0.1m/sec. Each entry in the fourth column is the product of the preceeding two entries from that row. The estimated total distance is the sum of the estimated subinterval distances.

Distance Approximations
SubintervalRight Endpoint Speed(m/sec)Subinterval Duration(sec)Subinterval Distance(m)
First63.210632
Second86.510865
Third95.010950
Fourth98.210982
Fifth99.310993
Sixth99.810998
Seventh99.910999
Eighth100.0101000
Ninth100.0101000
Tenth100.0101000
(Over)estimated Total Distance9419

Each of the calculated subinterval distances can also be interpreted as the area of a rectangle with base on the t-axis and whose height is determined by a value of the speed function. This is illustrated in the left figure below. The ten time subintervals correspond to the ten rectangles in the figure.

Ten Time Subintervals
Right EndpointsLeft Endpoints
We note that some of the rectangles protrude above the speed curve. The total extent of this protrusion is the magnitude of the error in our distance overestimate of 9419m dropped. The right figure above shows the rectangles that arise if we use the left endpoints or beginning speeds to approximate distance on each subinterval. The sum of the areas of these rectangles works out to be approximately 8418m, as you can verify for yourself. Obviously, the left endpoint approximation is an underestimate of the actual distance fallen. So at least now we know for sure that the actual distance is somewhere between 8418m and 9419m.

Accuracy is enhanced in this method by subdividing the 100 second time interval into shorter subintervals. The following graphs depict the rectangles built using right endpoints on each subinterval where the number of subintervals is 20, then 50, and then 100. Note that in the last picture, to total area covered by the rectangles is visibly close to the actual area under the curve.

Right Endpoint Approximations
# SubintervalsRectanglesApproximate Area/Distance
209229m
509097m
1009049m

It would be tedious in the extreme to calculate the rectangle sum for these 100 rectangles, but this is just the sort of thing at which calculators excel. I used the RSUM program on my TI-83 to calculate the area/distance approximations in the last column of the table.

And finally, in order to represent the exact area under the curve (ie the distance fallen during the 100 second time interval), we should take the limit of these rectangle sums (using right endpoints, say) as the number of rectangles goes to infinity. Makes sense geometrically and theoretically, yes, but how in the world to make such a calculation in practical terms?

Well, yes, calculus does offer a solution and we will get to it in Section 5.3 and 5.4. So stay tuned!! Just to show off, right now I will use a pencil and paper to calculate the exact area under the curve...hang on a second...

Yup, the exact area is 100(90 + 10e-10)m or about 9000.0454m.