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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
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
-1=0.2u-0.04g/2
u=-4m/s
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
d=u^2/2g=16/40=.4m
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