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Do NOT play beer pong with these guys...


Delhommey

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The day I make a behind the back shot on a bounce back I'll be happy.

Incidentally, the worst beating I ever took in beer pong was against two of the girls from the Maryland basketball team (Then the reigning national champs). We took our two shots, made one, then never shot again. Hell of a game.

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The day I make a behind the back shot on a bounce back I'll be happy.

Incidentally, the worst beating I ever took in beer pong was against two of the girls from the Maryland basketball team (Then the reigning national champs). We took our two shots, made one, then never shot again. Hell of a game.

Thats awesome!

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Cut to 10 years after college:

Person 1: No poo? I went to X college, too! Did we ever hang out at the same places?

Person 2: Probably, don't know. I spent 90% of my time videotaping me and my buds doing beer pong trick shots for 4 years. It was awesome.

Person 1: Oh. What are ya up to now?

Person 2: I'm in between apartments right now, saving money at the 'rents while my managerial career at the local movie theater takes off.

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Cut to 10 years after college:

Person 1: No poo? I went to X college, too! Did we ever hang out at the same places?

Person 2: Probably, don't know. I spent 90% of my time videotaping me and my buds doing beer pong trick shots for 4 years. It was awesome.

Person 1: Oh. What are ya up to now?

Person 2: I'm in between apartments right now, saving money at the 'rents while my managerial career at the local movie theater takes off.

lol

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Scrum's been doing some research it seems....

For example let’s look at the segment where the guy tosses the ball in the cup off of a moving skateboard. The beauty of all of this is that although the motion of the ball is three dimensional, the motion in each direction is independent of the other two and with a little basic physics we can calculate when and how fast the ball has to be thrown to make the shot. The cup appears to be about 8.0 feet or 2.4 meters away from the thrower. We can’t see the toss too well but we’ll assume he throws the ball up at about a 30 degree angle and the release point is about one meter above the level of the cup.

In the vertical direction we have a constant downward acceleration due to gravity. The motion is simply up and down and is described by the following equation:

Δy = v0 sinθ t – 1/2gt2

Where y = -1.0 m, v0 is the initial velocity, v0sinθ is the initial velocity component in the vertical direction, θ is the initial 30 degree angle of launch, g is the acceleration due to gravity (-9.8m/s2) and t is the time of flight.

In the horizontal direction (parallel to the floor) there’s a constant velocity once the ball is released since there is no horizontally directed force acting on the ball during its flight across the room. The motion equation is:

Δx = 2.4 m = v0cosθ t

And v0cosθ is the constant horizontal component of the velocity.

Solving these equations simultaneously we get v0 = 4.0 m/s, and the time of flight of the ball t = 0.70s.

Now let’s add in the lateral motion. In this direction the ball is simply moving along with the skateboard even after it’s thrown (Newton’s First Law). Since the ball is in flight for 0.70 s that means the thrower has to release the ball exactly that much time before the skateboard crosses the line of the cup. For a skateboard moving at 1.0 m/s that means he has to throw the ball 0.70 m before he arrives in front of the cup.

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