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

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     * 1.56m *
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