24 Feb '13 23:08

After the try at cone volume, let me see if I can make any headway with another long standing question:

How high does one need to throw a basket ball so it passes through the hoop without touching it?

Looking at basketball rules, the hoop diameter is 18 (inches) and the mens' basketball has a 29.5 inch circumference, meaning its diameter is 29.5/pi (inches)

To make the contact equation more simple for me to think about, I am going to "thicken" the basketball hoop wire to a 29.5 inch circumference and assume the basketball is a point. This will not change the answer but now at least my "thick" thing is static.

The path of any basketball is a parabola, for my first try let me assume it is the very simple parabola y = K.x^2, where K is positive. This turns our scenario upside down because this parabola is a U. The cross section of the thickened wire is a circle nestling in the bottom of the parabola We are doing to define X=0,Y=0 to be the point at the bottom of the parabola, and we are going to see if there is a viable solution where the circle touches the parabola at just one point (0,0), if we can find such a solution with a K small enough for the parabola not to hit the back of the hoop then we have proved that there is a way to throw the basket ball at hoop_height+ball_radius + d and (only just) not touch the hoop, where d is an infinitesimal amount.

So the equation of the circular cross section of the "thick" hoop (that the parabola has to avoid) is

x^2 + (Y-R)^2 = R^2

and the equation of the parabola is Y=Kx^2

So the points at which the circle and the parabola touch are solutions to:

x^2 + (Kx^2 - R)^2 = R^2

multiplying out and rearranging:

X^2 + K^2.X^4 -2KX^2 + R^2 = R^2

X^2 + K^2.X^4 -2KX^2 = 0

X^2(1 + K^2.X^2 - 2K) = 0

this obviously has a solution at 0. Because of symmetry the parabola will intertect the circle either at one point (0,0) or at 3 points, and it can only intersect at 3 points if (1 + (KX)^2 - 2K) = 0 has 2 real solutions

we can rearrange that to (KX)^2 = (2K - 1)

and we see that this will only have 2 real solutions if K is greater than 0.5. As the parabola "widens" with K, the solution least likely to hit the back of the hoop is K=0.5, i.e. the parabola Y=0.5*X^2

To get a handle on whether this parabola misses the back of the hoop let us calculate the Y value of the parabola when X = 18-29.5/pi which is the first point that the parabola is at an X value occupied by the back of the hoop.

so Y = 0.5*(18-29.5/pi)^2

Y = about 37.06

since the thickened hoop is only 9.4 inches or so in diameter, we have shown that with this parabola, we are well "below" the back of the hoop by the time the ball gets to an X value where it overlaps the hoop X.

Therefore I have shown the ball needs to be thrown to hoop_height+ball_radius+delta, where delta is infinitesimally small.

How high does one need to throw a basket ball so it passes through the hoop without touching it?

Looking at basketball rules, the hoop diameter is 18 (inches) and the mens' basketball has a 29.5 inch circumference, meaning its diameter is 29.5/pi (inches)

To make the contact equation more simple for me to think about, I am going to "thicken" the basketball hoop wire to a 29.5 inch circumference and assume the basketball is a point. This will not change the answer but now at least my "thick" thing is static.

The path of any basketball is a parabola, for my first try let me assume it is the very simple parabola y = K.x^2, where K is positive. This turns our scenario upside down because this parabola is a U. The cross section of the thickened wire is a circle nestling in the bottom of the parabola We are doing to define X=0,Y=0 to be the point at the bottom of the parabola, and we are going to see if there is a viable solution where the circle touches the parabola at just one point (0,0), if we can find such a solution with a K small enough for the parabola not to hit the back of the hoop then we have proved that there is a way to throw the basket ball at hoop_height+ball_radius + d and (only just) not touch the hoop, where d is an infinitesimal amount.

So the equation of the circular cross section of the "thick" hoop (that the parabola has to avoid) is

x^2 + (Y-R)^2 = R^2

and the equation of the parabola is Y=Kx^2

So the points at which the circle and the parabola touch are solutions to:

x^2 + (Kx^2 - R)^2 = R^2

multiplying out and rearranging:

X^2 + K^2.X^4 -2KX^2 + R^2 = R^2

X^2 + K^2.X^4 -2KX^2 = 0

X^2(1 + K^2.X^2 - 2K) = 0

this obviously has a solution at 0. Because of symmetry the parabola will intertect the circle either at one point (0,0) or at 3 points, and it can only intersect at 3 points if (1 + (KX)^2 - 2K) = 0 has 2 real solutions

we can rearrange that to (KX)^2 = (2K - 1)

and we see that this will only have 2 real solutions if K is greater than 0.5. As the parabola "widens" with K, the solution least likely to hit the back of the hoop is K=0.5, i.e. the parabola Y=0.5*X^2

To get a handle on whether this parabola misses the back of the hoop let us calculate the Y value of the parabola when X = 18-29.5/pi which is the first point that the parabola is at an X value occupied by the back of the hoop.

so Y = 0.5*(18-29.5/pi)^2

Y = about 37.06

since the thickened hoop is only 9.4 inches or so in diameter, we have shown that with this parabola, we are well "below" the back of the hoop by the time the ball gets to an X value where it overlaps the hoop X.

Therefore I have shown the ball needs to be thrown to hoop_height+ball_radius+delta, where delta is infinitesimally small.