A falling stone takes 0.15 seconds to travel past a window that is 1.0 meter tall. From what approximate height above the top of the window did the stone fall?

Question

A falling stone takes 0.15 seconds to travel past a window that is 1.0 meter tall. From what approximate height above the top of the window did the stone fall?

Answer 1

Meg,

 

If we divide the height of the window by the time it took for the stone to travel past we get the average velocity:

 

V_ave = (1 m)/(0.15 s) = 6.67 m/s

 

The stone was falling under the influence of gravity so it velocity increased as it moved between the top of the window and the bottom of the window.  In terms of it initial velocity at the top of the window and final velocity at the bottom we can write:

 

(V_i + V_f)/2 = V_ave = 6.67 m/s

 

and with the acceleration due to gravity we know that V_f = V_i + a·t = V_i + (9.8 m/s²)·(0.15 s) = V_i + 1.47 m/s

 

Let's rewrite our equation for average velocity to eliminate V_f :  V_f = 13.34 m/s - V_i

 

Combine this with our last equation:  13.34 m/s - V_i = V_i + 1.47 m/s

 

so 2·V_i = 13.34 m/s - 1.47 m/s = 11.90 m/s ⇒ V_i = 5.95 m/s.  This was the rock speed when it at the top of the window.

 

To find the height:  d = ½·at² and V_i = a·t, where t is the time that it took to fall from rest.  Eliminate t in the 1st equation to get: d = ½·a(V_i/a)² = V_i²/2a

 

Then d = (5.95 m/s)²/[2(9.8 m/s²)] = 1.8 m