Galileo applied his theory
of freely falling bodies to military problems such as "ballistic trajectories".
The useful abstraction here is the idea that, in a single motion, you can consider
separately the "horizontal" and "vertical" parts of that
motion. Gravity only influences the vertical part. It does not, in principle,
affect the horizontal part. In order to understand that abstraction, we will
investigate here some standard "pictures of motion."
These pictures are, of course,
graphs. They sometimes give a whole, qualitative view that it would take many
words (and equations) to convey. Suppose, for example, an object is launched
straight up from the ground. It goes up, then falls down. We can measure its
height at any instant of time and make a table, or we can draw a graph such
as the one below.
The horizontal
axis represents the time, and the vertical axis represents the height above
the ground of a projectile launched straight up (or down). There is no sidewise
motion, only vertical motion, but we "spread it out" over time so
that it is easier to visualize. This object was launched from height 10 meters
at an upward velocity of 25 meters per second.
Position and Velocity Portraits
The blue graph
is the graph of height as a function of time: the position portrait.
And the red graph is the graph of velocity as a function of time, the velocity
portrait. Now the notion of velocity will require some thought if you are
new to these things. It is close to your intuitive feeling associated with a
speedometer on a car, except that it may be negative as well as positive. You
saw that a body falling freely moves faster and faster as time goes by. The
velocity of the body describes just how quickly it is going at any instant of
time.
Since it varies
with time (decreases in this case) it is called a function of time. It is most
easily defined, using the terminology of Calculus, as a limit. Average velocity
over positive intervals of time is easy to understand: it is the change in position
divided by the (positive) change in time. But velocity (or instantaneous velocity
as it is sometimes called) is to be thought of as the number to which the average
velocities tend as the interval of time about a point in time gets smaller and
smaller.
Notice that the
velocity graph is a straight line with negative slope.
Question 1: What does that slope represent, in Galileo's language ?
The position
portrait is, however, a parabola. Now there is a visual relationship between
these graphs that we should point out. Consider any point on the blue curve
(position portrait). Imagine that a "tangent line" is drawn at any
that point. This means that the line just touches the graph at the point, and
does not cross the curve. This line will have a definite slope, and that slope
is interpreted as the instantaneous velocity of the projectile at the time represented
by the point.
For example,
at the instant the projectile stops rising and begins to fall (at approximately
2.55 seconds) the curve "flattens out" and the tangent line is horizontal;
it has slope 0. This means that at that instant the velocity is 0. Now notice
that the red velocity curve crosses the time axis (becomes 0) at just that point.
That is the simple relationship between the position and the velocity portrait.
The velocity portrait graphs the slopes of the tangent lines to the position
portrait.
There is another
fruitful relation between these portraits. Let's ask: "How high did the
projectile travel before it reached this point where it began to fall?."
We may answer the question by calculating the area of the triangle under the
red graph. And then of course add 10 meters, because that is where the projectile
started in this picture. When the velocity is positive, the distance travelled
from the starting point is interpreted as area under the velocity graph. We
will see why later, but it is worth thinking about. So how high did it go (roughly)
assuming the velocity became 0 after 2.55 seconds ?
In a very true
sense, the two basic problems of the Calculus are: to recover the velocity portrait
from the position portrait (the problem of the differential calculus), and inversely,
to recover a position portrait from the velocity portrait (the problem of the
integral calculus). So it is important to understand these things. We invite
you to experiment now with various values of position and velocity. Get a feeling
for what the two portraits are representing.
Exploration: Position and velocity portraits
On the left hand screen,
we represent what are called position and velocity portraits of some motions.
You may select the initial values of position and velocity using the scrollers:
The "position"
means the "height". Here, it may vary from 0 to 50. The "velocity"
means the "speed at any instant" Positive velocity means it is moving
upward, and negative velocity means it is moving downward as time increases.
When you press the:
button, you will see two graphs like the ones pictured above.
Question 2: At the time when the position portrait "flattens out", that is, becomes horizontal, what happens to thevelocity portrait? Why? What does it mean for the vertical motion for the position portrait to flatten out?
You may experiment with these
graphs in several ways. Sometimes, you will want to read points from the screen.
Do that by clicking the Trace check box. When you do that, the coordinates of
the mouse pointer will be drawn in the adjacent fields as you move the mouse
over the window.
Sometimes, you may want a
closer look at the screen. You may zoom-in on a window by clicking the button:
In. A pointing finger will appear, and you may use it to select a rectangle.
Do this. Click and hold the left mouse button down where you want the upper-left
corner of the rectangle to be. Still holding it down, drag the mouse to the
lower right corner, then release it. This will cause the contents of that rectangle
to fill the entire window. You may later undo this by clicking: Out.
In the blue window, we will
represent ballistic trajectories. The horizontal axis represents the displacement
of a "cannon ball" along the Earth, the "horizontal" displacement.
We could make this a measure of time because this horizontal displacement is
in fact proportional to time. That is a consequence of Galileo's law of inertia.
The horizontal motion is undisturbed (until the cannon ball hits something)
and so it continues uniformly, covering equal distances in equal times.
Not so the vertical motion.
The vertical velocity decreases at a steady rate. You calculated that rate in
an earlier exercise. In fact, as you saw in the velocity portraits on this page,
the graph of velocity is always a straight line.
Question 3: Calculate the slope of the velocity line. For this, you may click on points on the window, as described above. In fact, for more accuracy, you might like to zoom-in first. Is the slope the same no matter what initial velocity and height? Where have you seen this number (the slope) before? You may want to use the symbolic calculator to do the calculation. For that, click the book on the navigation bar at the top of the page.
That will take you to the cover page. Then choose symbolic calculator from the menu. The instructions there will tell you how to proceed.
Question 4: When a line that is not vertical is drawn on a graph and two pairs of points are selected with the property that the difference in x coordinates of the first pair is equal to the difference in x coordinates of the second pair, then we expect that the differences in y coordinates of these pairs will also be equal. This is a characteristic property of lines, and it captures Galileo's idea that equal changes in one variable lead to equal changes in the other. In your own words, why do you expect this to be true? There is no single "right answer" to this question. We are only asking you here to examine and describe some of your ideas about lines.
Now there are two ways to
prepare the cannon for firing. You may select the elevation of the cannon (its
angle measured in degrees with respect to the ground) by selecting the value
on the scroll bar:
And you may control the speed with which the cannon ball leaves the cannon (for example, by modifying the explosive force of the charge) by printing the new value in:
When you want to shoot the
cannon, press the: 
button. If you want to see the trajectory of the cannon ball, press the 
button. You may speed things up a little by making the 
larger (say, 0.1) but this will of course, decrease the accuracy.
Question 5: Try to hit the target by modifying the angle of elevation of the cannon only. Do not modify the speed yet. What angle sends the ball the greatest horizontal distance? What is that distance?
Question 6: Try to derive a formula for the distance traveled horizontally as it depends upon the speed, and the angle the cannon makes with the ground. Two factors come into play here. One is the constant horizontal speed (which depends of course on the angle of elevation of the cannon) and the other is the amount of time the cannon ball remains in the air. That depends also on the angle of elevation. If you developed a formula for the vertical motion of a freely falling body, you may calculate this time, and then you will be able to see what needs to be done to determine the distance traveled horizontally as it depends on the angle the cannon makes with the ground, and the initial speed of the cannon ball. If you succeed, then you will be able to predict how far the cannon ball will go before you fire it!
Now, modify the speed to
get a feeling for the shape of the trajectories of the cannon ball. If you want
to move the target, press the: 
button. After that, the cursor becomes a cross-hair. Press and hold the left
mouse button down as you drag the target to its new location. When you are satisfied,
release the left mouse button, and press the right mouse button. The location
of the target is reported on the screen below the button.
Question 7: If your objective is to make the cannon ball rise as high as it can for a given speed, what angle should you use?
Question 8: With the target at a reachable location for a given speed, in general, how many different angles will cause the cannon ball to hit it? If you wanted to program a computer to hit the target, you would need a formula, or an algorithm to tell it, given the position of the target, and the speed of the cannon ball, what angle to use to hit the target. This formula must take account of Galileo's laws. Can you find such a formula or algorithm?