 Posers and Puzzles

1. 03 Oct '11 08:15
How high does one need to throw a basket ball so it passes through the hoop without touching it?

For this one I have no solution yet. Assume that the center point of the ball follows a parabola and passes through the hoop the way they usually do, that is, downward. All relevant constants are considered known - where the ball starts its trajectory, the size of the ball and the size of the hoop.
2. 05 Oct '11 14:13
Originally posted by talzamir
How high does one need to throw a basket ball so it passes through the hoop without touching it?
In principle, just high enough to clear the hoop. The ball is smaller than the ring, so if it passes over the ring at the apex of its trajectory, it should also miss its sides going down.

Richard
3. 05 Oct '11 15:34
The problem is that if the trajectory is just barely high enough to nearest point of the hoop, it may well hit the farthest point. A basketball ring has a diameter of 18" or so, and a ball around 9". If the near and far point of the ring are at (0,-4.5) and (18, -4.5) or a hair's breadth lower, and the ball passes through (0,0) and that's the apex, the trajectory of the center point is y = ax^2 for some a < 0. If |a| is small the trajectory is flat and the ball will come with 4.5" of point (18, -4.5). If |a| is large, the ball will collide with the nearest point.. perhaps already on the way up... so I am not yet convinced. A further hurdle is that while the center point of the ball travels along a parabola, the path of the ball is not bound by two parabolas due to the ball being round rather than a point or a stick or a box. That is.. if the flight of the ball is recorded by a slow-exposure camera, the orange banana shape is not bound by y = ax^2 +/- r, but by something a bit more complex. What exactly the curves are .. well, knowing that would probably help quite a bit.
4. 05 Oct '11 20:56
Originally posted by talzamir
The problem is that if the trajectory is just barely high enough to nearest point of the hoop, it may well hit the farthest point. A basketball ring has a diameter of 18" or so, and a ball around 9". If the near and far point of the ring are at (0,-4.5) and (18, -4.5) or a hair's breadth lower, and the ball passes through (0,0) and that's the apex, the traj ...[text shortened]... complex. What exactly the curves are .. well, knowing that would probably help quite a bit.
All that matters is the trajectory. I play basketball and you can do this easily while standing underneath the hoop and throwing it just high enough to clear the rim.
5. 05 Oct '11 21:11
Originally posted by tomtom232
All that matters is the trajectory. I play basketball and you can do this easily while standing underneath the hoop and throwing it just high enough to clear the rim.
You could never complete a half-court shot while throwing the ball only high enough to clear the rim. The ball would hit the rim on the opposite side of the hoop.

At a minimum, the ball needs to fall a distance 2*ball_diamater faster than the time it takes the ball to travel horizontally the diameter of the hoop.
6. 05 Oct '11 21:32
Originally posted by forkedknight
You could never complete a half-court shot while throwing the ball only high enough to clear the rim. The ball would hit the rim on the opposite side of the hoop.

At a minimum, the ball needs to fall a distance 2*ball_diamater faster than the time it takes the ball to travel horizontally the diameter of the hoop.
He never said from half court... he said how do you need and if the trajectory and placement are correct then you only need to clear the rim.
7. 05 Oct '11 21:422 edits
I think there is a certain range of

Assuming the near edge of the rim is at (0,0) and the far edge at (18,0)
The ball has a diameter of 9 units.

Graph the following:

Radius around near edge that must be cleared:
y = sqrt(-x^2 + 4.5^2)

Radius around far edge that must be cleared:
y = -sqrt(-(x-18)^2 + 4.5^2)

I estimated the minimum angle the ball can take:
y = -0.11x^2 + 4.5

And the maximum angle the ball can take:
y = -0.035x^2 + 4.5

These are only visual estimates I made using graphing software, but anything outside of this approximate range of angles must be thrown higher than the rim to pass through the hoop w/o touching.
8. 05 Oct '11 21:44
Originally posted by tomtom232
He never said from half court... he said how do you need and if the trajectory and placement are correct then you only need to clear the rim.
Yes, but he meant using any given height and distance to the rim.
9. 05 Oct '11 23:431 edit
Certainly doable from half-court, at least in theory. Of course, you need a field cannon for it, an awesome aim, and probably an outdoor court. If the ball is launched to 200 feet in the air, it whooshes almost in a vertical line when it passes through the hoop, so hitting anywhere near the center of the hoop would suffice.

But.. to remove confusion, let's say that it's a three-pointer, so the ball starts at a distance of 22 feet from the basket, at a height of seven feet. The hoop is set at a height of 10 feet, a 9" diameter ball is used, and the hoop has a diameter of 18". NBA stars sink throws like that regularly in every game.. but how high at least does the ball need to go for the satisfying whooosh sound as it passes through the hoop without touching it?
10. 06 Oct '11 01:45
Originally posted by talzamir
Certainly doable from half-court, at least in theory. Of course, you need a field cannon for it, an awesome aim, and probably an outdoor court. If the ball is launched to 200 feet in the air, it whooshes almost in a vertical line when it passes through the hoop, so hitting anywhere near the center of the hoop would suffice.

But.. to remove confusion, let ...[text shortened]... ed to go for the satisfying whooosh sound as it passes through the hoop without touching it?
There aren't enough equations of motion for the defined parameters to arrive at a solution
11. 06 Oct '11 08:27
Okay, but what's missing? All the necessary data should be there.
* the location of the hoop and it's diameter are known.
* the location of the ball and it's diameter are known.
* the overall shape of the trajectory is known. A parabola that opens straight down.

Of all the parabolas that take the ball through the hoop, downward and without bouncing anywhere, it stands to reason that one could find an apex for which there is none that is lower, and it still makes the ball go whoosh. But which one? Shaquille and Kobe find it handily enough.

And that's what's so annoying about it. It looks clear and simple enough. But isn't.
12. 06 Oct '11 14:291 edit
Originally posted by talzamir
Okay, but what's missing? All the necessary data should be there.
* the location of the hoop and it's diameter are known.
* the location of the ball and it's diameter are known.
* the overall shape of the trajectory is known. A parabola that opens straight down.

Of all the parabolas that take the ball through the hoop, downward and without bouncing a ...[text shortened]... h.

And that's what's so annoying about it. It looks clear and simple enough. But isn't.
Using the kinematic equations the position "y" of the center of the ball at any time "t" in the "y" direction can be modeled as

^ y
|
|_____>x

y(t) = y*o + v_o*t + 1/2*g*t^2

Here are the knowns of this equation

y_o = initial height = H_o
g= acceleration due to gravity (assumed to remain constant)

So as it stands we have 1 equation and 3 unknows

y(t) = height of ball at time "t"
t = time
v_o=initial velocity in y direction

unfortunately it gets worse, the initial velocity has some magnitude and direction
so

v_0 = v*sin(A) (where "A" is an angle between the x&y axes, and v is the magnitude of the initial velocity)

the equation now looks like

y(t) = H_o + v*sin(A)*t + (1/2)*g*t^2 (1 eq, 4 unknows)

one of the unknowns can be eliminated by analyzing what happens in the x direction

Assuming the balls velocity in the x-direction remains constant

v*cos(A)*t = x(t)

You could then solve this equation for time "t", and eliminate the variable by substitution into the previous equation

y(t) = H_o + v*sin(A)*(x(t)/(v*cos(A))) + (1/2)*g*(x(t)/(v*cos(A)))^2

Now, the position parameters (x(t),y(t)) can be defined for the point where the ball is entering the rim

so you are left with 1 equation and 2 unknows ( the unknowns being: v, A)

one of which will have to be stipulated. ( considering the equation, the angle "A" is most economic to stipulate)

after the definition of the new parameter solve the equation for the other parameter

subtitute it into the original equation giving y(t) ( y as a function of time)

take the differential: d(y(t))/dt

constrain dy/dt = 0

solve for "t"

back sub into original equation, and "presto" you have your apex height

If all that was confusing just consider the standard form of the quadratic equation

y = ax^2 + bx +c

you'll need 3 coordinate pairs to solve for the coeficients a,b, & c; You have defined 2
13. 06 Oct '11 16:38
All that looks totally valid. I've generally used the equation

y = ax^2 + c ( a < 0)

for the trajectory, and moved the hoop so that the apex is on the y-axis. More exactly, the nearest and the farthest point are at some (d,0) and (d+18,0), d > 0.

So the problem becomes, to find the smallest value for c where

* ax^2 + c = 0 is true for some value of x in the range ]d, d + 18[
* the least distance between (d,0) and (t, at^2+c) > 9 for all values of t; and
* the least distance between (d+18,0) and (t, at^2+c) > 9 for all values of t.

That is, the ball passes through the hoop, doesn't impact with the near point, and doesn't impact with the far point.

The distance is pretty easy to calculate by using pythagoras.
Square roots fade nicely from it.
a or c could be replaced by using a number that depends on the point where the center point is at the same height as the hoop.
It seems logical to assume that the trajectory is so flat that ball almost touches both the nearest and the farthest point, so,

* ax^2 + c = 0 is true for some value of x in the range ]d, d + 18[
* distance between (d,0) and (t, at^2+c) = 9 for exactly one value of t
* distance between (d+18,0) and (t, at^2+c) = 9 for exactly one value of t.

..which is as far as I got. Various f'(?) = 0 solutions have so far failed.
14. 06 Oct '11 18:101 edit
Originally posted by talzamir
All that looks totally valid. I've generally used the equation

y = ax^2 + c ( a < 0)

for the trajectory, and moved the hoop so that the apex is on the y-axis. More exactly, the nearest and the farthest point are at some (d,0) and (d+18,0), d > 0.

So the problem becomes, to find the smallest value for c where

* ax^2 + c = 0 is true e value of t.

..which is as far as I got. Various f'(?) = 0 solutions have so far failed.
trying to move the coordinate axis around to depress the quadratic doesn't do any good

forget about the range of values; Lets just try to define 1 equation for the shortest distance to go into the hoop.

you chose to define the apex at x=0....ok

so the coordinate pairs of interest are now

(0,y_apex),(x1,y1),&(x2,y2) and you are trying to satisfy the equation

y= ax^2 + c

y1 and y2 are known, since the y axes now passes through the apex, c = y_apex

y(0) = c = y_apex

y= ax^2 + y_apex

to find the constant "a" we need just 1 more point, but surprise you now have no idea where either x1 or x2 lie relative to the apex

the best you can do is

x1 = x2-d

and you get the system

y1 = a(x2 - d)^2 + y_apex (1)

y2 = a(x2)^2 + y_apex (2)

once again 3 variables, only 2 equations.
15. 07 Oct '11 08:581 edit
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, so the closest point is

(u, sqrt(1-u^2)).

That point is on the curve.
Furthermore, a tangent of a circle and the radius are at a square angle towards each other. The direction of the line is dy / dx = sqrt(1-u^2)/u so the direction perpendicular to it is -dx/dy = -u/sqrt(1-u^2). So at that point, the differential of the curve is

y'(u) = 2au + b = -u/sqrt(1-u^2)

Thus, b = -u/(1-u^2) - 2au.

As (u,sqrt(1-u^2) is on the curve, a can be expressed in terms of u.

ax^2 + bx + c = y
au^2 + bu + c = sqrt(1-u^2)
a = [sqrt(1-u^2) - bu - c]/u^2

But b is known in terms of a and c, so

a = [sqrt(1-u^2) - (-u/(1-u^2) - 2au)u - c]/u^2

which solves as

a = [sqrt(1-u^2) - u^2/(sqrt(1-u^2) + c]/u^2
b = -u^2/(1-u^2) - 2[sqrt(1-u^2) - u^2/(sqrt(1-u^2) + c]/u

So a and b both depend only on u and c.

The curve passes exactly at the exact radius of 1 from (4,0) which allows u to be solved. That leaves a and b dependent on c only. That theoretically should allow the parabola to written as

y(x,c) = a(c) x^2 + b(c) x + c

For any given c, a(c) and b(c) can have a variable number of solutions, but that should give all the parabolas that pass the front end of the hoop at height c, for any c, so it gives all the solutions. One or more of which has the apex lower than all others. The most obvious way to find that would probably be to set

dy / dx = 0

though finding the x-coordinate of the apex at -b(c) / (2a(c)) could work too.

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