It takes 0.220 s for a dropped object to pass a window that is 1.14 m tall. From what height above the top of the window was the object released?

Answer 1

Let d be the distance from the drop point to the top of the window.
The time to reach the top is
t = sqrt(2d/g)
The time to reach the bottom is
t' = t + 0.220 = sqrt[2(d+1.14)/g]
= sqrt(2d/g) + 0.22

It's rather messy, but there you have one equation in the single unknown, d. Solve for d.

Answer 2

Let:
d1 = distance from drop to top of window
d2 = distance from drop to bottom of window
t1 = time from ball drop to top of window
t2 = time from ball drop to bottom of window
vi = initial velocity
g = acceleration due to gravity
Knowing the equation (d = vi*t + (a*t^2)/2)
d1 = (g*t1^2)/2
d2 = (g(t1+.22)^2)/2
d2 = d1 + 1.14
SO:
(g(t1+.22)^2)/2 = (g*t1^2)/2 + 1.14
Solve for t1.
Use t1 = sqrt(2*d1/g)
Solve for d1 and you will have the distance from the drop to the top of the window.