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Supercomputing Challenge

# Baseball With Degrees

Team: 40

School: JACKSON MIDDLE SCHOOL

Area of Science: Gravity

Interim: Team number: 40
School name: Jackson Middle School
Area of Science: Gravity
Title: Baseball with Degrees

Problem Definition:
If you shoot a baseball at different angles and at the same speed, what angle would have it land the farthest? Say you have a pitching machine and you shoot a baseball at several different angles, what angle would make it land the farthest?

Problem Solution:
We guessed about 45 degrees was going to be the farthest one, and it was. We tried 25, 35, and 45 degrees at about 50 and 60 MPH on the top wheel. Also the bottom wheel about 60 MPH on the tries we did.

Progress To date:
We had Java and NetBeans practice with Nick Bennett after school during November. Also, Randy Cabeen has been showing us formulas to figure out the distance of the pitching machine throwing a baseball. During December we traveled to Manzano High School and worked with Charlie Schlosser, the baseball pitching coach. He shot balls from the pitching machine, and we measured the distances at different angles.

Team members:
Ryan O’Rourke
Kevin Kortkamp
Marty Cabeen
Aundre Huynh
Scott Patrick

Mentors: Nick Bennett
Randy Cabeen
Charlie Schlosser

References

http://www.projectview.org/MathandBaseball/ScienceattheBallgame.htm
November 17, 2006

http://library.thinkquest.org/11902/physics/range.html
November 17, 2006

Physics/algebra/trig Eugene Hecht first edition, publisher?

Distance equations
vx0=x/t

y = vy0 t - .5 a t2

t = x / vx0

y = vy0 x / vx0 - .5 g (x / vx0)2
“Where did the g (gravitational constant) come from? Well, since the ball is moving through the air, gravity will be the force causing it to fall back down. Thus, in y = vy0 t - .5 a t2, the acceleration a towards the earth will be equal to gravitational constant g, which is about 9.8 meters per second per second (yes, two "per seconds". That simply means that every second, the velocity of the object increases by about 9.8 meters per second).
Don't be scared by the complexity of this equation! The trick is that when the ball hits the ground, you will be 0 (as well when the ball is originally hit). So the new equation will be:
0 = vy0 x / vx0 - .5 g (x / vx0)2
add .5 g (x / vx0) to both sides:
vy0 x / vx0 = .5 g (x / vx0)2
Divide both sides by (x / vx0):
vy0 = .5 g (x / vx0)
Multiply both sides by vx0:
vx0vy0 = .5 g x
And divide both sides by .5 g to get:
x = 2 vx0 vy0 / g “
Nick Bennett
Randy Cabeen
Charlie Schlosser

Team Members: