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(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
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