I drop Something from the roof past the window. I see that it took 0.13 s yo fall past the 2.0m y'all window. How far above the top of the window is the roof?
4.00 = (Vi+Vi+g*t)*t 4.00 = 0.26Vo+9.8*0.13^2 Vo = (4.00-9.8*0.13^2)/0.26 = 14.75 m/sec h = Vo^2/2g = 14.75^2/19.6 = 11.10 m
Past the window T = S/Vavg = S/(V + U)/2 = 2S/(U + gT + U) = 2S/(2U + gT) and 2UT + gT^2 = 2S; so that U = (2S - gT^2)/(2T) = (4 - 9.8*.13^2)/(2*.13) = 14.7 m/s atop the window. Then H = U^2/2g = 14.7^2/(2*9.8) = 11 m above the window top. ANS.
Let t = time for the something that U dropped to reach top of window. Then T = 0.13 + t = time for that which U dropped to reach bottom of window Let d = distance from roof to top of window Vt = speed of something dropped at top of window = gt = 9.81t Vb = speed of something dropped at bottom of window = gT - 1.2753 + 9.81t Vavg = avg speed of something dropped as it passes by window = (Vt+Vb)/2 Vavg = (9.81t+1.2753+9.81t)/2 = 19.62t/2 + 1.2753/2 = 9.81t + 0.63765 Vavg = distance/time = 2.00/0.13 = 15.3846 m/s 15.3846 = 9.81t + 0.63765 9.81t = 15.3846 - 0.63765 = 14.75 t = 14.75/9.81 = 1.503258 s d = 1/2gt? = (0.5)(9.81)(1.503258)? = 11.084 ≈ 11 m ANS
