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Grade: 12
        
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

Answers : (2)

Vikas TU
9517 Points
							
Let H be the height from the top to the window, then
h = H + 2 + 4 => H + 6............(1)
Now,
v^2 = 0^2 + 2gH
and
for window,
2 = (root(2gH))*0.2 + 0.5*g*(0.04)
on solving we get,
H = 80/2g => 4.05 meter
and from eqn.(1) we get,
h = H + 6 => 10.05 meter.
2 years ago
Sheharyaar
12 Points
							
Let u be x
v2= u2 + 2gh
so 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+gt
sq root(x2 + 40) = x + 10*0.2
                             = x + 2
squaring both sides,
x2+40 = x2 + 4 +4x
x2-x2 + 40 -4 = 4x
36 = 4x
 x=9
u = 9m/s
this initial velocity is the final velocity of the object from the top to the top part of windows
v = 9m/s
u=0 since the ball was dropped
g=10m/s2
v2-u2 = 2gh
v2-u2/2g=h
81 – 0 /2*10=h
h=81/20=4.05m
total 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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