help

Hello guest!

At t = 0 a ball is thrown straight upward from ground level at speed v0. At the same instant, a distance D above, a ball is thrown straight downward, also at speed v0.Where do the b***s collide (in terms of only D, v0, and g)?

Tonight came a new realization:
The b***s meet twice!

H = D - v₀t - (g/2)t²

(g/2)t² + v₀t + H - D = 0

  a          b         c

$t = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$t = {-v \pm \sqrt{v^2-2g(H-D)} \over g}$

H =  v_ot - (g/2)t²

(g/2)t² - vt + H = 0

   a        b     c

$t = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$t = {v \pm \sqrt{v^2-2gH} \over g}$

${\color{blue} {-v \pm \sqrt{v^2-2g(H-D)} \over g}}$= ${\color{blue}\frac { v \pm \sqrt{v^2-2gH}}{ g}}$ 

2v${\color{blue}\pm \sqrt{v^2-2gH}}$ = ${\color{blue}\pm\sqrt{v^2-2gH+2gD}}$

H must be calculated from this.
Maybe someone can help me.
For given sizes, the thing would be simple.

Greetings asinus :- )  !

Greeting asinus :- )   !