Results 46 to 60 of 105
Thread: The Arc on Joe's ball.

Re: The Arc on Joe's ball.
Haha it only takes preAlgebra to calculate. Just draw a triangle from where Flacco was when he threw the ball, to where Jacoby was when he caught the ball, and then draw a straight line back to where Flacco was when he threw it, and then finish connecting the dot. Then just do the Pythagorean theorem.
But it will be hard because it's hard to determine how many yards the ball sailed to the right.
EDIT: Oh, you mean including height and stuff. Oops.Last edited by PurpleApocalypse37; 01142013 at 04:23 PM.

01142013, 04:28 PM #47Rookie 24x7 Raven
 Join Date
 Jan 2013
 Posts
 14
Re: The Arc on Joe's ball.
This is the greatest and funniest thread I have ever seen. Love it.

01142013, 04:49 PM #48
Re: The Arc on Joe's ball.
I estimated the distance to the right as the distance from the hash to the sideline (the ball was thrown from and caught within a yard of two of each). Knowing that the has marks are 18.5 feet apart and the width of the field is just over 53 yards (2x+18.5=160), I estimated the short leg of the right triangle to be 24 yards. That's what gave me the 59 yards total.
Long days and pleasant nights.

01142013, 04:55 PM #49
Re: The Arc on Joe's ball.
Didn't the Ravens hire a numbers guru. Maybe he could chime in.

Re: The Arc on Joe's ball.

01142013, 05:21 PM #51
Re: The Arc on Joe's ball.
OMG Thread of the WEEK. LOL

01142013, 05:32 PM #52

01142013, 05:49 PM #53Regular 1st Stringer
 Join Date
 Apr 2012
 Posts
 845
Re: The Arc on Joe's ball.
So assuming the vertical displacement is negligible, we have a distance traveled of 49 meters in a time of 3.2 seconds.
We can calculate vertical velocity by using
0 = (initial velocity)(3.2) + 1/2 ( 9.8) (3.2^2)
This comes out to an initial vertical velocity of 15.68 m/s
We can then approximate max height using
0 = 15.68^2  2 (9.8)(max height)
This gives us a max height of ~ 12.544 meters or about 36 feet
Initial horizontal velocity can be calculated using
49 = x(3.2) + 1/2* .1 * (3.2^2)
This gives us an initial horizontal velocity of approximately 15.475 m/s
Using the calculated stuff here, we can use the following:
Integral from 0 to 3.2 of
sqrt((15.47  .1t)^2 + (15.68  9.8t)^2)
Which gives us an arc length of 74.551 meters, or approximately 81 yards.
Pretty awesome. Note these are rough figures, since we can't really be completely precise with the information we haveLast edited by CptJesus; 01152013 at 06:43 PM. Reason: Messed up integral

01142013, 05:50 PM #54
BaltimoreBoy  Do yourself a favor and email that to ESPN/sports science with emphasis on whether Moore could have made that play. That would be a sweet segment and I can't recall an episode where they focused on ball distance / height.
“Great minds discuss ideas. Average minds discuss events. Small minds discuss people.”
–Eleanor Roosevelt

01142013, 05:52 PM #55Regular 1st Stringer
 Join Date
 Apr 2012
 Posts
 845

01142013, 05:56 PM #56


01142013, 06:10 PM #58Regular 1st Stringer
 Join Date
 Sep 2011
 Posts
 498
Re: The Arc on Joe's ball.
This thread will end poverty and end all wars.

01142013, 06:10 PM #59Regular 1st Stringer
 Join Date
 Apr 2012
 Posts
 845

Bookmarks