CATAGORY: Mechanics QUESTION: A falling stone takes 0.30s to pass a window which measures 2.1m top to bottom. From what height did the stone fall?

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Lets first find the initial velocity of the falling stone as it appears at the top of the window. We will label the y height at the top of the window as y =2.1m and the bottom of the window as y =0m from the displacement formula (1) y = yo + vot -.5gt^2 Where g =9.8m/s^2. Substituting t =.3s y =0, yo =2.1m we have (2) 0m = 2.1m +vo(.3s) -.5(9.8m/s^2)(.3s)^2 Solving for vo (3) vo = (-2.1m +.5(9.8m/s^2)(.3s)^2)/.3s =-5.53 m/s This vo becomes the final velocity V for a fall from a height H to the top of the window. assuming the stone is simply dropped so its initial velocity is Vo = 0 we have from energy conservation (4) V^2 = 2gH Solving for H (5) H = V^2/2g = (-5.53)^2/2(9.8m/s^2) = 1.56m The stone was dropped from 1.53 m above the top of the window ********* * 1.56m * *********