The Physics of Beer Pong

Tossing a ping-pong ball into a beer cup? It takes more physics than you might think

These guys are pretty amazing. And the nonchalance with which they accomplish each trick shot adds a certain understated humor to this entertaining video. But though the guys seem to be developing a seemingly useless (if highly impressive) skill in their spare time, there’s quite a bit of complex science at play. In addition to being a highlight at any party, these are excellent demonstrations of two- and three-dimensional projectile motion, and with just a little bit of quantitative analysis the entire video would make a formidable project for an introductory level college physics class.

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.

We could do similar analyses for the other trick shots in the video, but I have a feeling that for the “trick shot masters” it’s more about the art than the science. That and a lot of practice!

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