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`        a stone is dropped from a top of a building of height ‘h’. an observer from inside the building clocks the time for the stone to pass down a window as 0.2s. if the height of the window is 2m, find the height of the building. the bottom of the window is at a height 4m from the ground.`
2 years ago

```							Let H be the height from the top to the window, thenh = H + 2 + 4 => H + 6............(1)Now,v^2 = 0^2 + 2gHandfor window,2 = (root(2gH))*0.2 + 0.5*g*(0.04)on solving we get,H = 80/2g => 4.05 meterand from eqn.(1) we get,h = H + 6 => 10.05 meter.
```
2 years ago
```							Let u be xv2= u2 + 2ghso v2 = x2 + 2gh = x2 + (2×10×2) = x2 + 40   if g=10m/s2  h of window is 2m. v = sq root(x2 + 40) v=u+gtsq root(x2 + 40) = x + 10*0.2                             = x + 2squaring both sides,x2+40 = x2 + 4 +4xx2-x2 + 40 -4 = 4x36 = 4x x=9u = 9m/sthis initial velocity is the final velocity of the object from the top to the top part of windowsv = 9m/su=0 since the ball was droppedg=10m/s2v2-u2 = 2ghv2-u2/2g=h81 – 0 /2*10=hh=81/20=4.05mtotal height of the building is 4.05  + height of window + height of tower from the lower part of the window= 4.05+ 2+4 = 10.05m
```
2 years ago
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