Let's start with Scriptor's clarification request. If the two balls
are deflated, then the answer to your question is yes.
Now, if we're talking about two fully inflated, one by one, side by
each, official basketballs, we need to consider the circumstances,
starting with a definition of 'diameter.'
"The length of a straight line through the center of an object from
side to side; width; thickness; as, the diameter of a tree or rock."
Now, we need to know the official diameter of a basketball hoop.
"Today's basket is an 18-inch-diameter (46 cm) metal rim, with a 15-
or 18-inch-long (38 or 46 cm), open-ended nylon net extending below
What is the diameter of a regulation basketball hoop? : Basketball
So we know how big the hole is, but what about the object(s) we want
to pass through it?
"The official basketball of both men's N.B.A. and N.C.A.A. leagues is
a size 7. This ball has a diameter of apporimately [sic] 9 inches, and
a total circumference close to 30 inches."
The world of basketball is full of geometry
We've got the facts, but the answer can't be derived from them so
easily. Here's my take on it. It is 'possible,' but not probable. One
could attach the said balls together precisely and, by hands-on
pressure, force the balls through the hoop.
One could attach the said balls together precisely and, by means of a
free-throw, connect with hoop, but without added pressure, no matter
how precise the throw, the balls would not pass though.
Perhaps Michael Jordan could make one of his amazing slam-dunks, and
if he's lucky, force the balls through.
Another scenario is that the balls are attached and pass through the
hoop vertically, as opposed to horizontally. In that case, there would
be no reason the two balls couldn't pass through easily, at least when
dropped through. To be shot through, it would depend on the skill of
the shooter, but it's possible.
I think the answer to your question, in theory, is yes. In practice,
barring a fluke, your answer is no.
diameter basketball +hoop
Clarification of Answer by
06 Feb 2004 16:32 PST
Hi again, yar,
In your 'clarify question' request you say, "... thrown basketballs
going thru the hoop at the same instant, side by side, without
touching each other or the hoop....no tricks"
The hoop = 18 inches in diameter
The ball = 9 inches in diameter
1 ball x 2 = 18 inches in diameter
There is no way two balls, thrown simultaneously, can fit
simultaneously into an equal space without touching each other or the
edge of the receptacle through which they need to pass.
As a matter of fact, the ONLY way the balls could go through the hoop
is if someone did exert downward pressure. It wouldn't be possible if
we were talking about rigid bowling balls or golf balls, whose shape
can't be altered. A basketball has a flexible exterior and is
air-filled. Thus, although its volume remains constant, its shape can
be altered under pressure.
If you take the time to watch the process in your mind's eye, you'll
see that the difficulties presented are virtually insurmountable.
Five yards to the left and eight feet out from the basket is
Meadowlark Lemon. Five yards to the right and eight feet out from the
basket is Michael Jordan. At exactly 7:43 pm, each player shoots, with
equal force, on the basket. Now, what's going to happen?
The two balls meet over the basket and bounce of each other. Neither,
much less both, ball passes through the basket.
The two balls were shot with just enough force that they drop to the
basket without touching, so no bouncing. They don't touch, and
assuming they are exactly equal in height, there is no way 18+ inches
can pass simultaneously through an 18 inch hole.
I'm not sure how I can better explain the conundrum, but I'm willing to try.