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 y_{2} is the final position and y_{1} 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 t^{2} term should be the acceleration divided by two. Using the parabolic fit, this gives a vertical acceleration of –11.54 m/s^{2}. True, this is *not* the expected value of –9.8 m/s^{2}, 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.

How Fast Was He Launched From the Cannon?

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: