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 A falling stone takes 0.2 seconds to fall past a window which is 1 m high. From how far above the top of the window was the stone dropped?

Please explain in detail

6 years ago


Answers : (2)


height of the window is 1m and it takes 0.2 seconds to pass, from this data we can find the initial velocity of the stone when it was at the top of window......

by using s=ut+at^2/2



let it travelled  distance d before passing through the window

so  v^2 =u^2+2ad

     v=-4,u=0(as the particle was dropped) and a=g



6 years ago
u=initial velocity of the stone
h=height above the window
s=height of the window
v0=initial velocity of the stone when it passes by the window
t=time taken to pass by the window
g=acceleration due to gravity (10m/s2)
The stone will begin with initial velocity 0 m/s.
Using 3rd equation of motion,
v02 = 2gh+u2
v0= \sqrt{2gh}
Now, using second equation of motion for motion of the stone when it passes by the window.
s=v0t+ ½ gt2
 \sqrt{2gh} t= s – ½ gt2
(Squaring both sides)
2ght= (s – ½ gt2)2
Substituting the given values,
2x10xh x 0.04 = (1 – ½ x10x 0.04 )2
0.8h = (0.64)2
So, h = 0.8 m
5 months ago

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