1. ## Physics again

1/2t^2+i*t+h=c

t=time passed
i=initial velocity
h=initial height
c=current height

I plugged in a few things, trying to find "t".

t=?
i=0 (falling, so 0)
h=10m
c=5m

And the answer I got was:
t^2 = -10

THAT'S IMPOSSIBLE!

What's wrong with the formula, because I checked the math w/ my teacher, my dad, etc. So it isn't me.

Any ideas?

2. heh, first, what the hell kind of notation uses 'i' for velocity? Second, the general formula is like this

y = y0 + v0t + 1/2 a t^2

y0 is original position
y is final position
v0 is the velocity
a is the acceleration
t is the change in time

A positive v0 implies moving in the positive direction. A positive a implies accelerating in the positive direction. There-in lies your problem, a was negative. Things do not accelerate away from the ground (after you release them, at least).

3. ## Sorry

I think it was you that I posted this to. I must have given you the wrong formula:

y=-1/2t^2+vt+i

where t =time passed
v=initial velocity
i=initial height
y=current height

I am sorry but I forgot to make 1/2 negative when I have it to u.