Basket Ball

Originally posted by talzamir
I tried it with this. Scaled the the image so that we have a parabola that passes between points (0,0) and (4,0) at a range of 1.

The equation of the parabola is

y = ax^2 + bx + c

or where convenient, x = t, y = at^2 + bt + c.

The points at the range of 1 from (0,0) are x^2 + y^2 = 1. Let u be the x-coordinate where the ball is closest to origo k too.

Thus, the number of equations is sufficient. Finding the solution, though. is tough.

This analysis baffles me...

first of all

(0,0) and (4,0) define a parabola; namely y= x^2 - 4x... Its max height is defined

Then you go on with this u substitution analysis where you intersecting the parabola with a unit circle???

you then seem to falsely conclude that perpendicular, to tangent, to the circle @ u is tangent to the parabola @ u with this equation??

y'(u) = 2au + b = -u/sqrt(1-u^2) (also your -dy/dx in terms of "u" should be positive here)

then further strange analysis thus follows from this point, which I can justify none of.

I mean, the parabola passes through the x-axis between the points (0,0) and (4,0). Those points represent the nearest and farthest point of the hoop. I got tired of having constants around so I scaled everything down 4.5 : 1, and fixed the origo at the hoop.

There is an infinite number of parabolas that pass through any two given points. For (0,0) and (4,0) y = a (x^2 - 4x) works just fine and the max height is -4a, with a some constant.. less than zero if we want a parabola that opens down.

I used the circles around the two points on the hoop. Saying that the parabola's nearest point is at range 1 from a given point is equivalent to saying that it touches a circle with that given point as center and radius 1.

No false conclusion there. If O is the center of a circle and P a point where the circle has a tangent drawn to it, that tangent is perpendicular to the straight line that passes through O and P.

It is possible I got a sign mixed up somewhere there. The point was simply to show that there are enough equations there to get an answer.. in theory anyhow. And that trying to get to that answer turns the nicely tame equation of a parabola into a really nasty monster. Basically doable, but in practice I have yet to get through it.

I hope this makes my thinking clearer? It could be a matter of a language barrier. Then again, it could be that my thinking was muddy to begin with.

Originally posted by talzamir
I mean, the parabola passes through the x-axis between the points (0,0) and (4,0). Those points represent the nearest and farthest point of the hoop. I got tired of having constants around so I scaled everything down 4.5 : 1, and fixed the origo at the hoop.

There is an infinite number of parabolas that pass through any two given points. For (0,0) and (4 ...[text shortened]... matter of a language barrier. Then again, it could be that my thinking was muddy to begin with.

Yeah, its probably mathematical barrier that I've never crossed, and that you most likely have. I was obviously totally misunderstanding you...I just hope you weren't snickering to yourself at my blissfull ignorance. 😳...😉

No snickering. I thought it was due to a language barrier on my side.

Don't you find it interesting how much easier it is to figure this out through practice shooting a ball than it is to solve the equation?


(I believe this will be misinterpreted because of my weak vocabulary but when I say "figure out by practice" I don't mean you figure out the numerical value that describes how high the ball goes but rather the practical value of height that you need to shoot the ball. I would say that both values are equal but one comes from an equation and I don't know if it has been explained how we know the other.)

Originally posted by tomtom232
Don't you find it interesting how much easier it is to figure this out through practice shooting a ball than it is to solve the equation?


(I believe this will be misinterpreted because of my weak vocabulary but when I say "figure out by practice" I don't mean you figure out the numerical value that describes how high the ball goes but rather the prac ...[text shortened]... ne comes from an equation and I don't know if it has been explained how we know the other.)

yeah, I agree...It's much eaiser to just do it!

But i would disagree eith the statement that the actual height is the same as what it "theoretically" should be.

Taking this to an actual basketball court would be fun too.

For me this has the same quality as the paper cone thing. A problem that looks nice and attractive, but given a bit of attention it turns into a rather hideous monster.

Dont forget also the "dipping" effects for giving the ball backspin as you toss it. This can also be affected more by a head-wind, causing the ball to dip more sharpley, making a more flatter shot capable of dipping into the rim "boneless",(without touching the rim!)

Originally posted by karoly aczel
Dont forget also the "dipping" effects for giving the ball backspin as you toss it. This can also be affected more by a head-wind, causing the ball to dip more sharpley, making a more flatter shot capable of dipping into the rim "boneless",(without touching the rim!)

Im pretty sure the problem is difficult enough...we don't need to add aerodynamic and angular effects. But if you think you have it..be my guest.

Not to mention a bit of spin that tends to have more effect when the velocity of the ball goes down, or gravity fields that don't uniformly point down etc. But even as is, the problem is rather tough.

A nearly equivalent problem without hoops or gravity or wind:

I dip a brush with a round tip with diameter d=1 in ink and paint parabola y = x^2 with it - that is, the center point of the brush follow that parabola. The paint turns black all points on the paper that are at a distance of up to 1 from the parabola. What are the two curves that bind the painted area?