It gives numbers on average flights of batted balls for each player. It's interesting to look at as a math application. I tried it out using formulas and didn't get the quite the same answer. However, the trajectory equations don't account for the air resistance encountered. And, of course, I might have just done the math wrong. More on this after I let you know how far off I was.
The categories were "Average Launch Speed", "Average Distance", "Average Velocity", "Average Launch Angle", and "Average Height". For example Evan Longoria (no relation to the actress) had:
- Average Launch Speed: 92.17 miles per hour
- Average Distance: 248.2 feet
- Average Launch Angle: 14.39 degrees
- Average Height: 46.06 feet
I assume Launch Speed could be found with a radar gun. Launch Angle perhaps by camera, although it seems like that would depend on where the camera is in relation to the camera. Ideally, the camera would be pointed perpendicularly to the ball's trajectory, I would think, but that wouldn't always be the case.
So, I wondered if I could compute what they had for Average Distance. I'm guessing that is how far the ball before it hits the ground. But what if Evan hits a line drive and is caught? It went a certain distance, but would have gone farther without the fielder there? Anyway, here we go.
First I figured I need to get its average speed into feet per second to match with the other categories.
92.17 miles per hour = 286,657.6 feet per hour = 135.183 feet per second
I then used the formula: y(t) = h + (vsinA)t -16t^2.
I'll assume an height of the ball when making contact with the bat to be 5.5 feet. I want to see how long it takes to hit the ground (y(t) = 0).
0 = 5.5 + 135.183sin(14.39)t - 16t^2
Using the quadratic formula, this game me two answers, the positive one being 2.25 seconds.
Then I used this to find how far it went with x(t) = v(cosA)t = 135.183(cos(14.39))2.25 = 294.62 feet
According to that website (http://m.mlb.com/player/446334/evan-longoria) the distance is only 248.2 feet.
I was ready to call this a big old fail. But, perhaps not. Like I mentioned before, I'm not sure how they figure balls that are caught before they land or balls that bounce off the outfield fence. And are those distances found by observation of where the ball seems to land? Air resistance slows down the ball quite a bit. They say that the Colorado Rockies in mile-high Denver is the easiest place to hit home runs because of its thin atmosphere. The math equations assume a vacuum, so the formula would give a greater distance.
So, maybe my math is all right. Regardless, it's a nice math application.
