A Flower Pot Falling Past a Window?

As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is L_w. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.





Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).





From what height h above the bottom of your window was the flower pot dropped?


Express your answer in terms of L_w, t, and g.

A Flower Pot Falling Past a Window?
When passes top of window, has fallen h. Has velocity √(2 g h) (v² = u² + 2 a s)





Takes time t to go from h to h + L_w, accelerating at g


L_w = √(2 g h) t + g t²/2 (s = u t + a t²/2)





h = (4L_w² - 4 g L_w t² + g²t^4)/(8 g t²)





Simplifying


h = (g t² - 2 L_w)²/(8 g t²)





UPDATE: if you know the answer, then why are you asking? And if I've got it wrong, be so kind as to point out the flaw in my working.





UPDATE 2: I just noticed h is from the bottom of the window. No matter. Now it falls a distance of h - L_w to have velocity √(2 g (h - L_w)) when it passes the top of the frame. It then accelerates a distance of L_w for t, leaving the second part unchanged. This just introduces an extra term in the numerator of the equation for L_w of 8 g L_w t^2 which changes the solution for h to


h = (4L_w² + 4 g L_w t² + g²t^4)/(8 g t²) and the simplification to


h = (g t² + 2 L_w)²/(8 g t²)