This isn't really breaking news, but there was this story of a guy that dropped his iPhone 4 out of a plane. The man was quite happy to find his iPhone unharmed. It was partially protected by Griffin case, but still this got me thinking. Was this a miracle or not so crazy?

Here are the details that might or might not be important: It was an iPhone 4. Here are the specs from Wikipedia. The phone was dropped from plane going 150 mph and flying at 1000 feet above the ground. The phone landed in leaves and pine needles. I am not sure how far it moved while landing. Maybe I could just guess that it was 3 cm.

What is terminal velocity? Well, when an object is falling in the atmosphere there are two forces that mostly govern the object's motion. Here is a diagram.

The gravitational force is essentially proportional to the object's mass (especially when near the surface of the Earth). The air resistance can be modeled as:

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Where: ρ is the density of the air. A is the cross sectional area of the object. In this case, it would depend on the orientation of the falling iphone. C is some drag coefficient that depends on the shape of the object (like flat versus pointed) And v is the velocity of the object. Oh, 1/2 is 1/2 or 0.5.

So, as the object falls, it increases its speed. With increased speed comes a greater air resistance force. This means that the net force will be smaller (since these two forces are in opposite directions). Eventually, the air resistance force will essentially be as large as the gravitational force and the net force will be zero (the zero vector). What happens to an object with zero net force? It travels at a constant velocity. So, this is terminal velocity.

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Using the above model for terminal velocity, I can write:

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What don't I know in the above expression? Really, I just don't know C or A because I don't know the orientation. How would this thing fall? I assume it would tumble. If it fell flat, I could calculate the terminal velocity.

Well, the Wikipedia page on air drag lists a coefficient of 2.1 for a smooth brick. How is an iPhone not a smooth brick? Essentially it is the same thing, right? There is no reference for that value, but I saw that same thing come up on a few other sites. Either it is an accurate value or everyone is using the wrong value based on Wikipedia. Why shouldn't I also use it? I assume that it takes into account any possible tumbling motion.

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Using that value for C and the bigger area (the screen side), I get a terminal velocity of 12.2 m/s or 27.2 mph. Just for fun, what if it was falling with it's smallest edge downward, this would give a terminal velocity of 42.8 m/s (95 mph). I suspect it would be closer to the flat-side terminal velocity (maybe around 20 m/s).Could it survive?

Now that I have an estimate for how fast it is moving before it hits the ground, I can get an estimate for its acceleration while colliding. Essentially this is similar to my Dangerous Jumping Calculator. One way to get the acceleration is to use the following kinematic equation:

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Note that this is a one dimensional equation and that *s* is the distance over which the object accelerates. In the case of a stopping iPhone, the final velocity (v2) would be zero. This means that the acceleration (using the terminal velocity as v1) would be:

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The acceleration would actually be positive since the displacement while stopping (s) is negative. Using a stopping distance of 3 cm and a terminal velocity of 20 m/s gives an acceleration of 6666 m/2 or 680 g's. If I use a terminal velocity of 12.2 m/s, the stopping acceleration goes down to 250 g's. That seems high.Comparing to dropping on a hard floor

How about something you might be able to relate to. Who hasn't dropped their iPhone on the floor? Me, that is who. You know why? Because I don't have an iPhone. I do have an iPod touch, isn't that really the same thing?

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In order to make a comparison, what height would I have to drop a phone over a hard floor so that it has a similar acceleration? This is a tough question because the answer depends on how much the floor compresses when it hits. I am just going to take a wild guess here. What if the compression (of both the phone and the floor) is about 1 mm? Here I can just use my dangerous jumping calculator to determine the starting height.

Here is my Wolfram Alpha jumping acceleration calculator.

Instead of building another one, let me put in a starting height of 1 meter and a stopping distance of 1 mm. This would give a stopping acceleration of 9790 m/2 or around 1000 g's. Just dropping from 0.5 meters would give an acceleration of around 500 g's. So, I don't think this event is so crazy.

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Warning: Just because I estimate that this acceleration is not out of line, I would not recommend dropping your phone from an airplane. First, it might hit someone. Second, you might break your phone. Also, I think it would be a bad idea to drop your phone on the cement floor. Oh, but wouldn't that be cool if you did drop it and you measured the acceleration?

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