The Physics of Dropping Out of a Plane in an Inflatable Ball

The MythBusters wanted to test whether someone could survive the drop from an airplane in one of those inflatable hamster balls. But dropping a ball from a plane is tricky---especially if you want it to land at a particular location. How about dropping it from a helicopter at a lower altitude? How high do you have to drop the ball so it reaches terminal velocity before hitting the ground? Let's find out.

Suppose you take a tennis ball and drop it onto the floor. You can model the motion of this tennis ball over a short distance by saying that there is only a gravitational force pulling down on it (that's not technically true, but true enough). With that simple model you could find the speed of the ball upon impact. This is what you do in an introductory physics course.

Now drop that ball from the top of a building and your model won't really work. There is another significant force on the ball: air resistance. You can feel this force when you stick your hand out the window of a moving car. The force pushing on your hand depends upon the following:

• The speed of the car (v).
• The size of your hand (A).
• The shape of your hand (C).
• The density of the air (ρ).

You can pretty much change most of these factors (except the density of air) and explore this air resistance force yourself. This air resistance can be modeled (usually) with the following expression:

Of course that's just the magnitude of the air force, the direction of this force is opposite the direction of the velocity. If you drop a sphere, then the area is the cross sectional area---so the area of a circle with the same radius. The shape of the object is included in the drag coefficient (C). For a sphere, C = 0.47 and for air, the density is about 1.2 kg/m³.

So, let's think about a ball falling from rest. Perhaps we can look at three key times during this fall:

• When the ball is released, it isn't moving at all so that it has a velocity of zero m/s. This means that the air resistance force is also zero. The only force on it is the gravitational force pulling down so that it accelerates down. Actually, because of the gravitational force the downward acceleration would be 9.8 m/s².
• A short while later, the ball is moving down with some velocity. This means that there are two forces acting on it---the downward gravitational force and the upward air resistance force. The result of these two forces is a net downward force that is smaller than just the gravitational force. The ball still accelerates down but with a acceleration smaller than 9.8 m/s².
• As the ball continues to increase in speed, the air resistance force increases. Eventually, the air resistance and gravitational force are approximately equal. The net force on the ball at this moment is zero Newtons so the ball stops increasing in speed. We call this final speed the terminal velocity.

If I set the magnitude of the air resistance force equal to the weight (which is what happens at terminal velocity), I can solve for the speed at which this happens.

The two important variables in this expression are the mass and the area (m and A). Increasing the mass increases the terminal velocity, but increasing the cross sectional area decreases the terminal velocity. Putting a human in a giant inflatable ball won't increase the mass very much but will have a huge impact on the area.

Now for the fun part. Let's find out how high you would need to drop something to make sure that it reaches terminal velocity before hitting the ground. This is fun because it's not so simple (simple things aren't fun). If you drop a ball with no air resistance (or negligible), then it has a constant acceleration and you can use kinematic equations or some other method to find the final speed. But when you include air resistance, the net force (and thus acceleration) changes when the velocity changes. This makes it tricky.

One way to solve a problem like this is with a numerical calculation. The basic idea of a numerical calculation is to break a problem with non-constant acceleration into many small steps. During each step I can approximate the motion as though it did indeed have a constant acceleration. Trust me, this works. Here is a more detailed example in case you want to learn more.

Here is a numerical calculation in python (on trinket.io) so that you can run this modle yourself. Notice also that I put the values at the top that you can change to run with different parameters (you should try changing these to see what happens---don't worry, you can't break it). Just click the "play" button to run it and then click on the "pencil" if you want to edit it.

Notice that this is the vertical speed vs. time for both a non-air resistance object and a ball. When the non-air resistance object gets to the ground, I set the velocity at zero m/s. Also, at the end I print the final velocity of the big ball as well as the terminal velocity.

You could of course just change the initial parameters until you just barely get a terminal velocity---but why do they hard work when you can get a computer to do it for you? Here is a similar program that plots the impact velocity as a function of starting heights. To create this, I am going to have to use a python function (quick tutorial on functions).

This is a plot of final velocity vs. starting height. Feel free to change the mass or radius of the falling ball. I already ran this code for you---if you really want to see it, just click the "pencil" to edit.

Now if you need to drop some object such that it reaches terminal velocity, you know how high you have to go. Go ahead and look up the mass and radius of a baseball or a basketball. Which one should you have to drop from a higher starting position? Guess and then try it.

Note: if you have a very high density object, you might need to get to large starting heights. In that case, the density of air and the gravitational fields would change. If you want an extreme example of this, check out the Red Bull Stratos Jump.