Sylvie asks:

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?

Back to catalog.

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 *

Back to catalog.