Mechanics

the particle at B moving downwards has -10m/s^2 acceleration, 0 initial velocity

the particle at A moving upwards has -10/m^2 acceleration, 50m/s initial velocity upwards.

let the distance between the meeting point and A = 200-m

let the distance between the meeting point and B = m

using the formula s = ut + 1/2at^2, form an equation for s using both 200-m and m and solve them simultaneously.

200 - m = 50(t) + 0.5(-10)t^2 ----- (1)

m = 0.5(-10)t^2 ------(2)

Sub (2) into (1):

200 - 0.5(-10)t^2 = 50(t) + 0.5(-10)t^2

200 = 50t

t= 4 seconds.

relative acceleration of both particles is zero this means that the relative velocity remains constant initial relative velocity is 50..........now time=(distance)/(speed)=4

for time t let the 2 objects move distances of h/x and h/y resply,such that h/x+h/y=h i.e x+y=xy.Then obj 1:h/x=50t^2-1/2(10)t^2 and OBJ 2:h/y=1/2(10)t^2. i.e y=h/5t^2..Substitute in x=y=xy. Then x=h/h-5t^2.substitute x in obj 1 displacement.SOLVE FOR TIME!!!