Jamaal Charles Passes to Himself - Real or Fake?

There's a video of Kansas City Chief running back Jamaal Charles in Pumas catching his own 45 yard pass. 

This is a pretty awesome feat for anyone, including NFL players.  Charles is one of the best backs in the NFL and (as a former track star) a very fast guy, but since we never saw the ball actually leave his hand, nor did we see it in flight until he catches it, could it have been faked?  Is it actually possible for him to have caught his own pass 45 yards down the field?  This isn't a feat made to look obviously fake, like Michael Vick throwing a ball out of the stadium, but is something that is in the realm of possibility.  Let’s look at the physics behind what he did.

There are two factors to look at in determining whether this feat is possible or not.  Can Charles realistically run that fast and can Charles throw that ball that high and far?

Let's see if our analysis can help determine if this is possible or not.  Warning, lots of math ahead!

The Run

Using a very unscientific method of repeatedly watching the video and using a stopwatch, I estimate Charles covered 45 yards in 5.8 seconds.  Most NFL running backs can cover 40 yards in under 5 seconds, so even if I'm off with my stopwatch, this is a very possible run for him.

In addition to seeing if this is realistic or not, we'll need to use this distance and time later.  Charles covers 45 yards, which is equivalent to 41.148 meters.  He covers this in roughly 5.8 seconds.  A quick division tells us his speed is:

41.148m/5.8s = 7.1 m/s

Putting that into Imperial numbers, Charles is running at nearly 16 miles per hour.  Certainly fast, but for an NFL running back over 45 yards, very doable.

The Pass

Figuring out whether this pass is possible is a little more complicated, since we could theorize the angle he threw the ball, the velocity in his hand, and various other factors, but we're going to simplify this problem by using vector physics.  We will boil down Charles' pass into two components and look at them individually to see if it’s a realistic throw.

1. How hard must he throw the ball upwards?  - How high will the ball travel?
2. How hard must he throw the ball forwards?  - How far will the ball travel?

Putting both of these factors together, and including the force of gravity, we can build a picture of the throw that looks something like this:

Vector diagram of what a pass might look like

How Hard Must He Throw the Ball Upwards?

Since we know it took Jamaal Charles roughly 5.8 seconds to run those 45 yards, he clearly must throw the ball high enough to stay airborne for those 5.8 seconds.

In physics, the position of anything in a line can be determined by its initial position, its initial velocity, its acceleration and the time since the start of measurement.  At this point, we're only concerned about how the ball moves in a vertical motion, not yet in its horizontal motion.  We know how long the ball stays in the air (5.8 seconds) and we know what the acceleration factor due to gravity is.

First, we have to figure out what the initial velocity in the vertical direction is.  Then, we can find out how high the ball gets before it comes back down to earth.

In more specific terms, the formula for determining the height of an object given an initial velocity and acceleration is:

position at time t = (initial position) + (initial velocity * time) + (1/2 * acceleration * time²)

Or in physics parlance:

y(t) = y₀ + v₀(t) + (1/2)gt²

In this formula, the "0" represents initial position or initial velocity and "t" represents the number of seconds.

Initial and Final Position

We will call the initial position of the ball 0.  The final position of the ball is also going to be 0, since Charles throws and catches the ball from roughly the same vertical position (height off the ground). This will make our calculation much easier.

Initial Velocity

We don't know how fast Charles is launching the ball vertically, so we'll leave this in as an unknown variable (v₀) which we'll try to figure out.

Acceleration

Once the ball is in the air, the only acceleration in a vertical direction is due to earth's gravity.  Scientists have calculated this figure to be about 9.81 m/s².  Since the earth is pulling downward on the ball, we represent this as a negative number.

Time

Again, Charles had the ball in the air for 5.8 seconds according to my admittedly unscientific method of watching the video again and again, and measuring with a hand timed stopwatch.  That number will be "t" in our equation.

Crunching the Numbers

Putting our numbers into our original equation, we now get the following new equation to find how hard Charles threw the ball in the air:

0 = 0 + v₀(5.8) + (1/2)(-9.8)(5.8)²

Simplifying:

0 = v₀(5.8) + -164.836

165.0042 = v₀(5.8)

v₀ = 28.42 m/s

Using these numbers, the ball must be launched with a vertical speed of 28.42 m/s.  Converting this number to a more familiar (to Americans) number, he's throwing the ball straight up in the air over 63 miles per hour.

How High Does the Ball Go?

The related question we can now answer is how high upwards does the ball travel before it comes back down.  Since the ball is in the air for 5.8 seconds, we know it is on its way up for 2.9 of those seconds, and on its way back down for the other 2.9 seconds.  Using the velocity number (v₀ = 28.42 m/s)we calculated in the previous equation, we can now change the time to 2.9 seconds to see how high the ball gets before it comes back down.

y = 0 + (28.42)(2.9) + (1/2)(-9.8)(2.9)²

y = 82.418 - 41.209

y = 41.209 m

The ball must go in the air over 41 meters, or about 135.2 feet, over 10 stories high!  That's pretty high, and surely a normal guy like me can't throw the ball that high, but could a professional football player?

How Hard Must He Throw the Ball Forwards?

This calculation is a little easier to work with, since we don't have to fight against gravity in a horizontal direction.  We're really only fighting the effects of the wind, which (for theoretical purposes), we'll simply ignore.  Charles must propel the ball in a horizontal direction 41.148 meters in 5.8 seconds.  This now becomes a simple distance = rate * time calculation and the equation becomes:

distance = rate * time

41.148m = v₀ * 5.8

v₀ = 7.1 m/s

Shockingly (or perhaps not), the exact same speed at which Charles has to run!  Given a little bit of thought, this is a somewhat obvious conclusion.  Both the ball and Charles start and end at the same point, so it's clear to reason he has to throw the ball with the same horizontal velocity as he's running.

Using Those Numbers to Look at the Pass More Closely

These numbers don't tell us the full story of course.  We also will want to know the angle he had to throw the ball to achieve these numbers, and ultimately, how hard did he have to chuck it?  Remember, the forward and upward velocities on the ball only give us part of the answer.  Finding the angle which he threw it gives us a better idea.

We can recreate our vector diagram with actual numbers this time.  Since we have some actual numbers to work with, we can draw our vector diagram a little bit more to scale:

More accurate depiction of Jamaal Charles' throw

Finding the Angle of the Pass

Finding the angle of the pass is relatively easy if you know trigonometry and have a calculator handy.  At this point, if you haven't learned trigonometry, it can get a little complicated, but bear with me here.  I can reorganize the triangle to look like this:

A right triangle

The angle θ, (theta,) is part of a right triangle.  The sides we know are opposite (28.42) and adjacent (7.1) to this angle.  We can use the tangent function, commonly abbreviated tan) to determine what the angle actually is.  This will require a little bit of calculator work, but the function will look like this:

Tan(θ) = 28.42/7.1

Tan(θ) = 4.0028

At this point, we know the tangent of this angle = 4.0028.  In order to find the angle itself, we have to take the inverse tangent of this number to find the angle, basically working backwards to find a value.

θ = Tan^-1(4.0028)

For this calculation, we need to punch it into our calculator:

θ = 75.9 degrees

Jamaal Charles would have to put the ball up nearly 76 degrees from the horizontal!

Finding the Velocity of the Pass

Finally, we want to find how hard Charles actually threw the ball, not just in its respective horizontal and vertical components.  Luckily for us, the triangle we're working with is a right triangle, so we can simply apply the Pythagorean Theorem to get the hypotenuse, which in this case represents the total velocity the ball was thrown at.

7.1² + 28.42² = v²

50.41 + 807.6964 = v²

858.1064 = v²

v = 29.29 m/s

This pass would have to have left Jamaal Charles' hand at 29.29 m/s or 65.5 miles per hour!  As we saw, most of that was going straight up, since he had to toss the ball at a pretty steep angle.

(One other way of calculating the velocity of the pass is also using sine or cosine to find the hypotenuse)

Conclusion

Using the numbers given, we conclude in order for Jamaal Charles to throw this pass in such a manner that he can run under it and catch it, the pass would have to have:

1. Been thrown at a 75.9 degree angle
2. At 65.5 miles per hour (29.29 m/s)
3. which would have soared to a height of 135.2 feet (41.209 m)
4. and landed 45 yards (41.148 m) from where he threw the ball.

Still, this doesn't answer the question: Is this fake? The proposed numbers speak for themselves, but without context, it's hard to know.  According to the bastion of truth that is the internet, a quarterback can throw the ball at over 60 mph for sure.  However, they are usually aiming for a target and not trying to leave it in the air for that long.  Charles, on the other hand, got a little bit of a running start and can chuck it as hard as he can in the air and try to run under it.

What about the actual angle of the pass?  Using an incredibly unscientific screen shot of the video, I've overlaid what appears to be the angle of his arm after just after he lets go of the ball.  See for yourself:

Certainly not an easy feat for anyone, but if anyone can do it, it's a professional football player.  Therefore I fully believe it's possible.

However, we will have to take most of this on faith because 1) we can't see the ball for most of the throw 2) the ball has to travel over 130 feet straight up in the air and 3) a RB is throwing the ball as fast as a QB.

As always, feel free to leave comments.  Please keep them civil though.