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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 mLet the body fall through the height of the tower in t seconds. From equation of motion we have sn=u+a2(2n−1) So, we have total distance travelled in last 2 seconds of fall as 40=st+s(t−1) ⇒ 40=[0+g2(2t−1)]+[0+g2(2t−3)] ⇒ 40=g2(4t−4)⇒ 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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