Posted by **Matt** on Friday, December 16, 2011 at 12:57am.

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?

- Physics -
**drwls**, Friday, December 16, 2011 at 1:44am
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.

- Physics -
**Sabrina**, Sunday, September 4, 2016 at 5:47pm
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.

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