Here's how I did it, you have to set up the problem into multiple parts first. Draw a picture and it makes sense. You're trying to solve it all in one equation and that's not possible because you have to solve for multiple variables, thus multiple equations are necessary. Here's a picture:
The various equations are (assuming constant acceleration):
1.
average Velocity across window = dX/dT = 1.90m/0.420s = 4.524m/s
2.
average Velocity across window = (V1+V2)/2
Solving for V1 we get (V1+V2)/2 = 4.524m/s
V1 + V2 = 9.048m/s
V1 = 9.048m/s  V2
3.Then solving for acceleration across 1.90m we get:
average Acceleration across window = dV/dT = (V2V1)/0.420s
V2 = V1 + at = V1 + (9.8m/s^2 * 0.420s) = V1 + 4.116m/s
V2 = V1 + 4.116m/s
Combining these last two equations we get:
V1 = 9.048m/s  V2
V2 = V1 + 4.116m/s

V1 = 9.048m/s  (V1+4.116m/s)
V1 = 9.048m/s  V1  4.116m/s
2V1 = 9.048m/s  4.116m/s = 4.932m/s
V1 = 2.466m/s is velocity at top of lower window.
4. Pluging into V=at for velocity over constant acceleration we get:
2.466m/s = 9.8m/s^2 * t
t = 0.252s time to fall to top of lower window
5. Pluging into d=1/2at^2 for distance over time w/constant acceleration we get:
d = 0.5at^2 = 0.5 * 9.8m/s^2 * 0.252s^2
d = 0.310m (don't round off numbers until final step)
Last edited by DannoXYZ; 092605 at 06:33 PM.
