I used this to figure it out: https://www.youtube.com/watch?v=4SHqHFWYg-s&t=99s
Simple physics models a parabola
Distance:
s = s_0 + v_0×t + 0.5×a×t^2
use a = -9.81 m/s^2
velocity:
v = ds/dt = v_0 + a t
these can be vector equations, so you measure distance, then height. The Tangent to the parabola should be heading almost downwards, say at angle of 60 degrees to the horizontal. (There is no arc length reparameterization of the parabola, so tangents arent so great here).
So to cut a long story short, you need a vector with the xz as cos(60) = 0.5 and the y as sin(60)=0.866
I.e. v = Vector3(0.5, 0.866, 0.0)
to get an arbitrary vector to the basket in into this simply get
var v_planar = (basket_pos -player_pos)
v_planar.y =0.0
v_planar =v_planar.normalized()*0.5
v_planar.y = 0.866
then scale by the required height, i.e. 3m higher than the players final throwing hand position
so from v= v_0+ at , v .y=0 = v_0.y -9.81*t, at max height, so
v_0.y=9.81*t
v_0.x = (0.5 /0.866)× v_0.y
So go at 60 degrees, the players arm extended might be only 1m lower than the hoop, and if the players are 7+ft tall then they can almost dunk.
Look up the old 2D game “2 tanks on a hill”.
The v_0.y initial y velocity sets the height, and time of flight, because only the y axis has acceleration.
Theres a solution here … i think ‘tg’ is ‘tan’ short for tangent …
tan a = sin a /cos a