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Question

A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of the tower is
(g=10 m/s2)


A
90 m
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B
45 m
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C
55 m
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D
60 m
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Solution

The correct option is B 45 m

Let the body fall through the height of the tower in t seconds.
From equation of motion we have
sn=u+a2(2n1)
So, we have total distance travelled in last 2 seconds of fall as
40=st+s(t1)
40=[0+g2(2t1)]+[0+g2(2t3)]
40=g2(4t4) t=3 sec
Thus, distance travelled in 3 sec will be the height of the tower which is given by
h=ut+12gt2=0+12×10×32
h=45 m


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