Question

ok, i have 2 balls, one thrown up vertically at an unknown velocity, and one dropped at the same time, from a building of an unknown hight(H). when they meet, the speed of the dropped ball is twice that of the thrown ball. How do i find the hight of the collision, in terms of H?

Answer 1

Since for the first ball, with constant aceleration h1=-1/2gt^2+vt, and for the second ball h2=-1/2gt^2+H, vt=H and t=H/v in the point of colision.
puttin in the formula, we get: --------h=-1/2g(H/v)^2+H--------

Answer 2

Can you use v also or not?

Answer 3

(2v1)^2 =2g*y2 (1)
v1^2 -vo^2=-2g*y1 (2)
(1) +(2)
5*v1^2 - vo^2 =2g(y2 - y1) , y2 = H - y1
we get y2=.....

Answer 4

we get y1=.....