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Question

A body is dropped form the top of a tower. It falls through 40 m during the last two seconds of its fall.The height of tower in m is (g=10msâˆ’2)

A
45 m
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B
50 m
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C
60 m
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D
80 m
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Solution

The correct option is A 45 mIn the case of free fall displacement of the motion is S=12gt2 let assume nth second = tn(n-2)th second = tn−2given that (tn−tn−2)=2...(1) Displacement in last two seconds,S(n,n−1)th=12g(tn−tn−2)2 putting values in this equation ⟹40=12×10×(tn−tn−2)2⟹(tn−tn−2)(tn+tn−2)=8⟹(tn+tn−2)=4...(2)by solving eq 1 and eq 2 we get the values of ⟹tn=3⟹tn−2=1hight of the building S=12gt2⟹S=12×10×32⟹S=45m

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