Understanding Calculus

e-Book for $4


  Table of Contents

  1. Why Study
  2. Numbers
  3. Functions
  4. The Derivative
  5. Differentiation
  6. Applications
  7. Free Falling
  8. Understanding
  9. Derivative
  10. Integration
  11. Understanding
  12. Differentials

  Inverse Functions
  Applications of
  Sine and Cosine
  Sine Function
  Sine Function -
  Differentiation and
  Oscillatory Motion
  Mean Value
  Taylor Series
  More Taylor Series


+ Share page with friends
Your Name:
Friend Emails:
Your Email - optional:
CHAPTER 7 - Study of Free Falling Bodies

Chapter 7 - Study of Free Falling Bodies

Section 7.2 - Understanding Free Fall Motion

Having  laid down the conceptual basis of what velocity, acceleration and forces are, we can now study the motion of free falling bodies on earth. An object falls to earth because of the gravitational force of attraction that the earth experiences for the object. What then is gravity? From Newton’s law of gravitation, the force attracting two bodies is given by:

To derive this, think of a unit mass of 1 kg separated a distance from a larger mass, M. The gravitational force is the force of attraction the larger mass expresses for the unit mass and vice-versa. Observation confirms that the gravitational force is proportional to mass the larger body and decreases with the square of the distance separating them. The reason it is distance squared and not just directly related to the distance is because masses are 3-dimensional. In 3-dimensional space, properties are related to the projected areas as opposed to 2-dimensional geometry that are dependent on the length only. Therefore, the gravitational force between the two masses is:

If the unit mass was replaced by a mass m, the force of attraction would be m times the amount it was with the unit mass.

G replaces “c” as a gravitational constant determined from experiments. The gravitational force of earth acting on a body of mass located near the surface of the earth is then:

Since most of our free falling bodies occur near the surface of the earth, we can take d, to be the radius of the earth. Substituting the known values in along with the value for G, reduces the equation to:

Since force equal ma, we have:

This important result tells us that the acceleration of a body of any mass is 9.8 meters per second per second near the surface of the earth. The gravitational force acting on a body near the surface of the earth would be its mass, m, times the constant acceleration, 9.8. For those who have a ground to hold them up this does not mean much but for a free-falling body in air its acceleration as it falls toward the earth will be a constant , regardless of its mass. A body of twice the mass will be pulled in by twice the force, but the acceleration due to the force of gravity remains the same.

This may sound a bit confusing but just remember any body will fall to the earth with a constant acceleration, independent of its mass. In terms of particle motion,  mass means nothing for a falling body! While the gravitational force increases with mass, the acceleration remains the same.

We can now write the acceleration function for a falling body near the earth’s surface as:

Since acceleration as a function of time is by definition the derivative of the velocity function with respect to time then what we have is the same as:

The derivative of the velocity function (acceleration) is 9.8. Since we know that the derivative of any function is then we can easily find the velocity function since we know what its derivative is.

The process of finding a function, given its derivative is known as anti-differentiation. In this case 9.8 can also be written as . We see that:

Therefore n equals 1. Consequently the anti-derivative of is . This should hopefully be obvious since the derivative of is just . We now know that velocity as a function of time is:

whose derivative with respect to time is:

The graph of the velocity function is a linearly  increasing function with constant rate of change or slope

This graph of the velocity function gives us the objects velocity as any time t, assuming that air-resistance is negligible. For example at t= 10 seconds, the object’s velocity is:

At t= 94 seconds its velocity is:

Or almost 30,000 km/ hr. Due to air-resistance no object reaches such high velocities. Remember the greater the height it is dropped from the more time it has to increase its velocity or accelerate before it slams into the earth.

Now how do we find the distance function or the distance covered from the point of dropping the object. From the definition of velocity we know that:

Since , we have n=1 so its anti-derivative will be where the derivative of is ct = 9.8t. The solution is then:

The derivative of this function is the velocity function or:

We can graph the distance function . The graph gives us the vertical distance traveled from where it was dropped at any time t.

Clearly as time, t,  increases, the rate at which distance is being covered is very great . For example between t= 0 s  to t =5s, the object has covered totally.

Or the object has covered 112.5 meters in the first five seconds of its free-fall. However from t = 20 s to t = 25 s, the object has covered:

The object has covered more than a kilometer during this five second interval!! This should make sense because initially the body’s velocity is small and thus does not cover much distance over a time interval ;. However, after some time its velocity has increased ( look at the graph of the velocity function graph), such that over a same interval , the object covers a greater distance. Remember constant acceleration means that the velocity is increasing linearly with time and distance increases with the half square of time.

Next section -> Section 7.3 - Initial Conditions for Motion


© Copyright - UnderstandingCalculus.com
Web Design by Online-Web-Software.com
Developers of Event Registration Software