MATH 164, Spring 2001 Due Date: Name(s):
Honors Project 7: Ballistics with Air Resistance
In MATH 163, we found that if -- neglecting air resistance -- a projectile of mass m is fired vertically
upward then its height y above the ground at any time is governed by the equation my = -mg where g
is the force due to gravity. It followed that v = y (t) = -gt + y0, and that y = y(t) = -1
2 gt2
+ v0t + y0.
Thus, if we use metric units -- so g = 9.8 m/sec2
-- then v = -4.9t + v0. A little reflection on this formula
for velocity might lead you to question the assumption that we are neglecting air resistance: as t increases
without bound, v = -4.9t + v0 increases without bound, and this is impossible. On one hand, practical
experience illustrates that objects can not fall at ever increasing speed; there is, in fact, a speed dubbed
terminal velocity beyond which a projectile cannot travel. On the other hand, if we accept the fact that no
object can move beyond the speed of light then this formula leads to a contradiction. In this project, we
investigate the affects of air resistance.
One way to incorporate air resistance into our discussion of the motion of a projectile, is to use the equation
my = -mg - cy (1)
where the term cy represents drag due to air resistance. Observe that cy is proportional to y : the larger
the velocity the greater the air resistance. The constant c is known as the drag coefficient; c depends on the
shape the projectile: the larger c the greater the air resistance.
To solve this equation, we treat it as a first order differential equation in v = y : we may rewrite it
mv = -mg - v
and solve it for v by separation of variables; and having expressed v = x (t) as a function of t, we can
integrate to find y = y(t).
Tasks
1. Type the following command lines into Maple in the order in which they are listed. The affect of these
commands is solve equation (1) assuming that m = 1.0, c = 0.2, g = 9.8, y0 = 0.0, and v0 = 128.
> # Honors Project 7: Ballistics with Air Resistance
> restart: with(DEtools):
> m := 1.0; c := 0.2; g := 9.8; y0 := 0.0; v0 := 128;
> y2 := diff(y1(t),t);
> dsolve(m*diff(y1(t),t) = -m*g-c*y1(t),y1(0)=v0, y1(t));
> y1 := unapply(rhs(%),t);
> dsolve(diff(y(t),t) = y1(t),y(0)=y0,y(t));
> y := unapply(rhs(%),t);
2. Continue by typing the following command line into Maple. The affect of this command is to plot the
solution to equation (1) for t  [0, 10].
> t0 := 0.0; t1 := 10.0; plot(y(t),t=t0..t1); y(t1);
3. By using trial-and-error, change the value of t1 in the last line you typed until you obtain the value of
t1 greater than t0 for which y(t1) is within 2 decimal places of 0.
4. Observe that the curve you get is not symmetric. Explain why it is not.
5. Find when the projectile achieves its maximum altitude, and find the maximum altitude.
6. Find the terminal velocity of the projectile.