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• Kinematics & Rotational Motion
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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

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### 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
-1=0.2u-0.04g/2
u=-4m/s
let it travelled  distance d before passing through the window
v=-4,u=0(as the particle was dropped) and a=g
d=u^2/2g=16/40=.4m


6 years ago
										Let,u=initial velocity of the stoneh=height above the windows=height of the windowv0=initial velocity of the stone when it passes by the windowt=time taken to pass by the windowg=acceleration due to gravity (10m/s2) The stone will begin with initial velocity 0 m/s.Using 3rd equation of motion,v02 = 2gh+u2v0= $\sqrt{2gh}$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)2ght2 = (s – ½ gt2)2Substituting the given values,2x10xh x 0.04 = (1 – ½ x10x 0.04 )20.8h = (0.64)2So, h = 0.8 m

5 months ago

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