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.

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) ... b) V-ground^2 - Vb^2 = 2g Hb V-ground^2 = 2g (h_ total) = 2g Hb + Vb^2 2g ( h_ total) = 2g Hb + Vb^2 ( h_ total) = Hb + Vb^2/2g ... c) From 1 (Vb) = (L/t) + 0.5gt The bottom of window is at a height Hb above the ground V-ground^2 - Vb^2 = 2g Hb V-ground^2 = 2g Hb + Vb^2 Substituting for Vb, we get the required equation.

Hey here is your answer