Free Fall, or “freely falling objects”

We are studying free fall at this point in the course because it is an easily accessible example of constant acceleration. If we can assume two things:

• The falling object is near the surface of the earth or some other very large object, and

• If there is no air resistance

Then the object will experience a constant acceleration as it falls.

As always, when we make simplifying assumptions, we should check them.

What does “near the surface of the earth” mean? For our purposes, let’s just assume that we can be within 1/1000th of the earth’s radius and we’ll be OK. (The math for this is a little beyond this point in the course, but it’s quite doable.) The earth has a radius of approximately 6400 km, so as long as we’re not dropping anything from more than 6.4 km, or about 4 miles above the surface of the earth, we can assume that gravity is constant enough for our purposes.

The “no air resistance” assumption is not quite as easy. The air resistance that an object experiences depends on a number of factors, including the speed of the object, the size of the object, and the shape of the object. So we have to make a judgement call. For something like a tennis ball being dropped from a couple meters, air resistance will not be significant. However, drop that same tennis ball from a couple hundred meters, and air resistance will become quite large. In this case, we cannot assume the acceleration is constant and we would need a bit more math to understand the motion.

If we can make those two assumptions — near earth’s surface, and no air resistance — then we can assume the acceleration is constant and caused only by gravity. In this case, that assumption becomes math:

A couple of notes about this equation before we go on:

• The negative sign: Our usual frame of reference says “up is positive".” Gravity works in a downward direction, so we generally assign a negative to the acceleration. In rare cases where the frame of reference says “down is positive,” gravity would be positive.

• The location of the negative sign: This one confuses lots of beginning students. “g” — the symbol for the acceleration of gravity — is a magnitude and therefore is always positive. (Review the vectors side quest if necessary.) The direction (the negative sign) sits outside the magnitude. If you see “g” by itself, it is positive.

We’re going to take the constant acceleration equations that were presented in Module 1.5 and make two changes to them:

• First, we’re going to replace “a” in the equations with “-g” as explained above.

• Second, because free fall is vertical motion, we’re going to replace “x” with “y” and we’re going to put a y-subscript on the velocity. It’s important to keep horizontal motion and vertical motion separate, as we will see in the next unit. Don’t shortcut the process and just call all your position variables “x.” If you do that, learning motion in more than one dimension will be difficult.

When we make those two changes, we get a set of “custom” constant acceleration equations that can be used for free fall:

If necessary, jump back to 1.5 and verify that these are exactly the same equations we saw before, but with the two changes described above.

“Unwritten rules” of free fall: One of the confusing things about free fall problems is that they seem to be lacking enough information. Here are some things to look for:

• Zero position:

  • An object that starts its journey at ground level — such as a football kickoff from field height — has an initial position equal to zero meters.

  • An object that ends its journey at ground level — such as a rock dropped from a tower — has a final position equal to zero meters.

  • It is quite possible for an object to have both initial and final positions equal to zero. Again, think of a football kickoff that is not caught, but hits the field.

• Zero velocity:

  • If an object is thrown up into the air, when it reaches its maximum height, it will stop for an instant before going back down. Setting the final velocity to zero will let you solve for the maximum height.

  • If an object is dropped from a height — not thrown upwards or downwards, but dropped — the initial velocity is zero m/s.

  • When an object is dropped, hits the ground and stops, the “final velocity” as far as a physics problem is concerned is NOT zero. The problem wants you to calculate the velocity the instant before the object hits the ground. (Remember that downward velocities are usually negative.)

To do:

• Watch, and if possible, try this free fall lab

• From OpenStax Physics section 2.7, try 41, 43, 45, 47, 49, 55