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 = vot - (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 :- ) 
!
