    Question

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

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

## The correct option is B 45 mLet the body fall through the height of tower in t seconds. We know that distance covered the body (Dn) at a particular second n is given by the relation Dn=u+a2(2n−1) As the body is at rest initially ⇒u=0 ∴Dn=a2(2n−1) Let t be the time taken by the body to reach ground. Distance travelled in last 2 seconds =40 m (given) Total distance travelled in last 2 seconds of fall can also be expressed as D=Dt+D(t−1) =[0+g2(2t−1)]+[0+g2{2(t−1)−1}] =g2(2t−1)+g2(2t−3)=g2(4t−4) =102×4(t−1) ⇒40=20(t−1) t=2+1=3 s Distance travelled in t seconds is s=ut+12at2=0+12×10×32=45 m  Suggest Corrections  1      Similar questions  Related Videos   Free Fall
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