if acceleration is -9.8, initial velocity is 10, initial position of a point is +0.5m. then when does it pass y=0 and what is the velocity at that time
You left out some dimensions in your problem that are important. I will asuume you meant 10 m/s for the initial velocity Vo, and -9.8 m/s^2 for the acceleration.
It passes zero when
y = 0.5 + 10t -4.9t^2 = 0
Take the positive root for the answer.
Using the quadratic equation,
t = (-1/9.8)(-10 -sqrt (100 + 9.8))
= 2.09 s
The velocity at that time is Vo + at
= 10 -9.8*2.09 = -10.48 m/s
You could also use conservation of energy to get that answer (but not the sign). The potential energy difference (per mass) at y=0, 9.8*0.5 = 4.9 J/kg, equals the increase in V^2/2 (K.E. per mass), which becomes
V'^2/2 = Vo^2/2 + 4.9 = 54.9
V'^2 = 109.8
|V'| = 10.48 (absolute value)