The trailer recently dropped for Jackass Forever, which arrives in October—the next in the series of Jackass movies in which Johnny Knoxville, Steve-O, and the rest do some pretty dumb stuff. I mean, these guys aren't exactly young anymore—but that’s not stopping them. Sure, it might be funny, but I just keep thinking about how much they could get hurt.
Since they have already done the stunts, we might as well make the most of them. Let’s go with the one in the trailer that shows Knoxville being shot out of cannon wearing a pair of wings as though he could fly. (Hint: He can't. But don’t worry, he lands in a lake.) Yes, it's sort of silly to shoot yourself out of a cannon, but it gives us some great physics questions to answer.
I'm going to start off with some data. I can use video analysis to get position and time data for Knoxville in each frame of the video. There are several ways to do this, but I really like Tracker Video Analysis. (Plus, it's free.)
Of course, there are some issues to deal with. I need to determine the size of an object in the background and be able to keep track of the location of a fixed origin, even though the camera both pans and zooms during Knoxville’s launch from the cannon. Tracker Video Analysis can help with the origin, but I need to know the size of an object we can see in the video. My guess is that the cannon they use is the same one used by Dave "The Bullet" Smith—a popular human cannonball act. It looks the same and it has a listed length of 35 feet, so I'm just going to go with that.
After moving and adjusting the reference frame, I can mark the location of Knoxville to get all the data. Here is a plot of his horizontal-position as a function of time.
I also have a plot of his vertical position as a function of time.
These two graphs will be very useful as we answer some pressing physics questions.
I think this is a fair question, because there's obviously lots of fake stuff online. (If you want a more detailed guide to finding fake videos, I wrote one here.) In this case, we can calculate whether the jackass in fact moves like a typical projectile.
If you launch an object off the ground, it will essentially just have one force acting on it—the downward-pulling gravitational force. This gravitational force is equal to the produce of the object's mass (m) and the gravitational field (g = 9.8 Newtons per kilogram). Since this is the only force, the object should have a vertical acceleration equal to -9.8 meters per second squared. Without a horizontal force, it should have a constant horizontal velocity.
Now we can check these two things.
If you look at the x-position data above, you can see that it appears to be a linear function, since the x-velocity is defined as:
But with a position-time graph, the change in position divided by the change in time is also the slope. From that, we can see two things. First, the horizontal motion does have a constant velocity. Second, the value of this horizontal velocity is 10.09 meters per second. (That's 22.6 miles per hour.)
What about the vertical motion? Since there is a constant acceleration, the vertical position should agree with the following kinematic equation (where y2 is the final position and y1 is the starting position):
The important thing is that this is the equation of a parabola. Looking back at the vertical position data from the video, it at least seems fairly parabolic. Even better, the coefficient in front of the t2 term should be the acceleration divided by two. Using the parabolic fit, this gives a vertical acceleration of –11.54 m/s2. True, this is not the expected value of –9.8 m/s2, but it's really close. (I suspect that my scale for the length of the cannon could be off by a little bit.)
Neither the x nor y-motions of Knoxville disagree with the expected physics. Does it mean the video is real? Nope—it could still be faked, but personally I think that it is indeed real. I mean, doing stupid stunts is the whole point of a movie like this.
Right when Knoxville leaves the cannon, he is moving in both the horizontal (x) and vertical (y) directions. We already know his horizontal velocity, so we would just need the vertical component of velocity.
However, there's a way to get the total velocity (we call that the magnitude of the velocity vector) just by using the launch angle. Looking at the video and using the protractor tool on Tracker Video Analysis, it seems like the cannon is angled 52 degrees above the horizontal. Since the horizontal and vertical velocities are perpendicular, I can draw the following right triangle:
With this being a right triangle, I can use the cosine as the ratio of the adjacent side (vx) to the hypotenuse (v total). But I know vx and the angle—so, there you go. That puts the total launch velocity at 17.7 m/s (39.6 mph). Yeah, that's pretty fast. It’s slower than a professionally-pitched baseball, but faster than you can run. This launch velocity will be useful to answer some other questions.
The trailer doesn't show Knoxville’s whole motion after being shot from the cannon, but that's OK. We can use our projectile motion equations to solve for this distance.
The key to any projectile motion problem is that the horizontal and vertical motions are independent, except for the time. That means that I can look at this projectile-human and just consider his vertical position and vertical velocity. I can then use this total time for the horizontal motion and find out where the dude hits the water.
Let's start with the vertical direction. If I set the water level to be y = 0 meters, then I need to approximate the starting y-position. Based on the video, it looks like he leaves the cannon about 9 meters above the water. I can get the initial vertical velocity using the launch angle and the magnitude of the velocity. (See, I told you it would be useful.) With that, I get a y-velocity of 13.9 m/s. Going back to the kinematic equation for constant acceleration (the vertical acceleration is -g) and using a final y-position of 0 meters (at the water), I get:
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It's not a completely trivial equation to solve. The problem is that there is a constant term, a term with t, and then a term with t2. Really, the only way to get a solution is with the quadratic equation. I'll spare you the details, but that method results in two times for the motion. They are –0.54 seconds and 3.4 seconds. You can't really have a negative travel time, so that means we will go with a time of 3.4 seconds.
Next, I can use this time and the constant horizontal velocity to find the horizontal distance traveled. Let's use a starting x-position of zero meters. That means we get the following:
With the values for time and x-velocity, this gives a landing distance of 37 meters (121.4 feet). Hopefully, the lake is bigger than that or Knoxville is not going to have a good time.
For the person in the cannon, they would have to go from a speed of 0 m/s to 17.7 m/s in just the distance of the cannon tube. Would that kill you? Well, it obviously didn't in this case—but let's calculate the force pushing Knoxville forward. I will start with a diagram.
Since we are dealing with a human going from position 1 (at the bottom of the tube) to position 2 (leaving the tube), this problem can be best addressed using the work-energy principle. This says that the work done on an object is equal to a force (from the cannon) multiplied by the distance that its applied (in this case L).
The work done on an object changes that object's energy. There will be two types of energy here—the gravitational potential energy (U) that depends on the height and the kinetic energy (K) that depends on the speed.
I'm going to set the y-value at position 1 equal to 0 meters. Of course, the velocity at position 1 is also zero. Putting this all together, the work-energy equation becomes:
I can get a y-value at position 2 using the length of the cannon (6 meters for shooting part) and the launch angle (52 degrees).
Putting this all together and using a Johnny-mass of 82 kilograms (180 pounds), I get a push-force of 2,774 Newtons. If I divide this force by his mass, I get his average acceleration at approximately 33.8 m/s2.
We often like to express accelerations in terms of "g's" where 1 g = 9.8 m/s2. That puts this cannon launch at 3.5 g's. That's not too bad. It’s about the acceleration of a human in a rocket, or maybe that feeling in the dip during a fast rollercoaster ride. (Just so you know, NASA has lots of data about the human tolerance of g-forces).
So there you have it: The Jackass Forever stunts might be silly and childish, but the physics is still real and even fun.
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