As you look out of your dorm window, a flower pot suddenly falls past. The pot i

Question

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 Lw. 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 Lw, t, and g.

Explanation / Answer

One of the equations of motions is s = ut - 1/2gt^2, Substituting for u = v-at, the equation becomes s = vt - 1/2at^2.
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In the problem instead of using LW, let us use L alone.
h is the height from the bottom of the window to the position where from which the pot was dropped
L = (Vb)t - 0.5gt^2 or
(Vb) = (L/t) + 0.5gt-----------------------------------...

Also (Vb) ^2 =2gh,

Hence
2gh = [(L/t) + 0.5gt] ^2
h = [(L/t) + 0.5gt] ^2/ 2g. This is the needed equaation.

This can be modified as
h = (L^2/2gt^2) + (L/2) + (gt^2/8)
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