physics and football by: ian ramdeen, chris, david, evan harrison, and brent richardson
TRANSCRIPT
![Page 1: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/1.jpg)
Physics and FootballBY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson
![Page 2: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/2.jpg)
Football
American football is obviously popular in the U.S., especially in the South.
Did you know that physics can also be found in football such as a football being a projectile, momentum of a player, force applied in a tackle, and forces used in a kick?
![Page 3: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/3.jpg)
Projectile in football
• A projectile is any object that can be thrown.
• In football quarterbacks throw the football for several miles at quick speeds. If a quarterback wants to throw a long pass, then he is going to have to throw at the right angle and right velocity.
![Page 4: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/4.jpg)
Practice Problem
If a quarterback throws a football at 20m/s and at an angle of 60 degrees, how far did he throw the ball?
Formula: R=Vo^2 sin (2 theta)/g
Theta=60 degrees V=20m/s G=9.8m/s^2
Solve for R
Solution: R=20^2 sin (2*60)/9.8
400sin(120)/(9.8)=35.348 meters
![Page 5: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/5.jpg)
Physics Problem
If the same qb throws a football at 25m/s and at an angle of 50 degrees, how far did he throw the ball?
Solution: R=25^2sin2(50)/9.8
63 yards. Wow! What an impressive throw! He gets my MVP vote.
![Page 6: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/6.jpg)
Momentum
Everyone knows that momentum is a measure that is equal to mass times velocity.
This applies to football in this manner because football players of different sizes are running at fast velocities.
![Page 7: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/7.jpg)
Physics Problem
If a 125 kg (275 lb) lineman is running down the field at 9.8 m/s, what is his momentum.
Formula: p=mv
M=125 kg v=9.8 m/s
Find p (momentum)
Solution: p=125 (9.8 m/s)
=1225 kg-m/s
![Page 8: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/8.jpg)
New Problem
If this same lineman runs at 7.5 m/s, what is his momentum?
Solution: 125(7.5)=937.5 kg-m/s
![Page 9: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/9.jpg)
Tacking and Impact
• One of the biggest concepts of physics applied to football is force. A force of a tackle can have huge impacts. If the impact of the tackle is huge, that can lead to a serious injury. If a running back is running with a huge momentum, a sudden tackle can come as a huge shock.
• (Rolando McClain gave this player a huge shock with this strong tackle).
![Page 10: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/10.jpg)
Physics Problem
• If a running back has a momentum of 950 kg-m/s, and the tackle occurs in .6 seconds, what would the force of the tackle be?
• Formula: F=impulse/t
• Impulse (Momentum but written as impulse in this case): 950 kg-m/s
• t=.6 seconds
• Find the Force
• Solution: F=(950 kg-m/s)/ (.6seconds)=1583 N
![Page 11: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/11.jpg)
New Problem
If McClain tackles former Auburn running back Ben Tate in .4 seconds, and his momentum is 1000 kg-m/s, what is the force?
Solution: F=1000/.4=2500N
![Page 12: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/12.jpg)
Physics and kicking
• Physics especially applies to the instance when a placekicker is trying to make a field goal. He has to make adjustments in order to make the attempt. He has to kick the ball at the right angle and has to kick the ball with a good force. He also needs to kick the ball well, so a good acceleration will enable the ball to go through the goal post. A good time while the ball is in the air enables the ball to go through the goal post.
![Page 13: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/13.jpg)
Physics Problem
• A place kicker must kick a football from a point 40 m from the goal and clear the crossbar 3.05 m high, and the ball leaves the ground with a speed of 25 m/s at an angle of 50 degrees to the horizontal at an acceleration of 9.8 m/s. How long does the ball stay in the air?
• Formula: D=ViT+(1/2)A(T^2)
• Given Vi=25m/s A=9.8 m/s D=40m T=?
• 40=25(T)+1/2(9.8)(T^2)
• 4.9(T^2)+25T-40=0 (multiply each side by 10)
• 40=25(T)+4.9(T^2) =-250 +/- the square root of 140900 divided by 98 seconds.
![Page 14: Physics and Football BY: Ian Ramdeen, Chris, David, Evan Harrison, and Brent Richardson](https://reader035.vdocument.in/reader035/viewer/2022062303/5518d3f6550346b31f8b5e17/html5/thumbnails/14.jpg)
References
http://www.lcse.umn.edu/speces/labs/catapult/index.html
http://entertainment.howstuffworks.com/physics-of-football3.htm
http://answers.yahoo.com/question/index?qid=20080930092439AA1Sgc8