A stone is dropped from height h. Simulataneously another stone is thrown up from ground with such a velocity that it can reach a height of 4h. Find the time when two stones cross each other.

the second ball can reach a height of 4h.    so, at 4h its velocity =0      v²=u²+2as    so, 0 =u² -2g*4h  so, u=sqrt(80h)
i have taken upward direction to be positive.
for the dropped stone, u=0    and for the thrown stone, u=sqrt(80h)
let they cross at time t    and the dropped ball covers distance/displacement x , then the thrown ball covers h-x.
so, for the dropped ball,  -x=0-(1/2)g*t²      and the second ball, h-x = [sqrt(80h)]*t -(1/2)gt²       sobtracting the second equation from the first one, we get h=[sqrt(80h)]*t      so, t=sqrt(h/80)

v2=root(8gh)