Dale Basler (@Basler) at Lab Out Loud found this great ad for Farmer's Insurance.
I think Dale did a good job at pointing out one correct solution to the projectile pig problem. He also points out that the picture seems to call s the distance traveled and uses units of m/s. Why? And here is my first point. This is a very nice drawing, meaning it looks professional. However, it is wrong. Why not take the extra hour to email Dale Basler (or a high school physics student for that matter) and say "hey, we are doing this ad with projectile motion. Would it be ok to do it like this?" But no. They didn't even think to check this. Physics doesn't need to make sense, it just needs to look complicated.
More projectile analysis
Rather than just leave this post as a rant, let me see if I can add some value to what Dale started. Question: are the trajectories shown parabolas? Using Tracker Video I can get x-y position data from the image. I will assume that it is to scale and I will put the range of the shortest flung pig at 85 meters.
Is it a parabola? Here is a quadratic fit from Tracker.
So, this first one does not look very parabolic. Maybe they were taking into account air resistance. I mean, if the pig is launched with a speed of 42 m/s I would suspect that is near the terminal velocity of a pig.
Ok, if there is no air resistance should the trajectory (x-y) be a parabola? Yes. This isn't as simple as it looks. For projectile motion (with no air resistance), there is no acceleration in the x-direction and the acceleration in the y-direction is -g. This gives the following two equations (that have the same time):
Now, I need to remove the time (t). Let me solve the x-equation for t and plug that into the y-equation. This give me:
Just to save time, let me shift the axes so that x0 and y0 are at the origin. This gives:
And there you go - quadratic equation. Why don't introductory students look at trajectory equations instead of y vs. time equations? Because you don't get the acceleration so easily. Here the term in front of the x2 is not (1/2)g like it is with the time-equation. Ok, I could find the value for g here, but I won't. I won't because the trajectory doesn't even look close to a parabola.
But what about the other two pig-shots? They look much more quadratic-like. Here is the farthest flung pig:
If this pig was indeed shot at 48 m/s with an incline of 70° (and there is no air resistance) then I can relate the fitting parameter (a = -0.012) to the trajectory equation. The initial x-velocity would be:
Now I can solve for g:
Ok, that is not 9.8 m/s2, but much closer than I thought it would be.
What should this really look like?
Again, ignoring air resistance the trajectories of these three little pigs should look like this:
I will leave the calculations with air resistance as a homework assignment.
Update:
Everyone says its a cow and not pig. Cow, pig - they are both spheres to me.
