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).

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).

Question:

If the bottom of your window is a height hb above the ground, what is the velocity vground of the pot as it hits the ground? You may introduce the new variable vb, the speed at the bottom of the window, defined by

vb=Lwt+gt2.

Express your answer in terms of some or all of the variables hb, Lw, t, vb, and g.

The correct answer is

vground = (t²+lw²+(2t³+2t²)×g×Lw+(t⁴+2t³+t²) ) ∧ 0.5 = vb² +2g×vbt+1/2gt²

Explanation:

We have from the relation

v = u + gt, S = ut + 1/2 gt², v² = u² + 2gS

in this case S = hb, u = vb=Lwt+gt2.  and v = vground

therefore v² = (Lwt+gt²)² + 2 × g × hb

= (Lwt+gt²)² + 2 × g ×  (Lwt+gt²)×t + 1/2 gt² = vb² +2g×vbt+1/2gt²

vground = (t²+lw²+(2t³+2t²)×g×Lw+(t⁴+2t³+t²) ) ∧ 0.5