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Question

# A body falls from rest under gravity and travels half of its total path in the last second. Find the time of fall. (Take g=10 ms−2)

A
2.8 s
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B
3.4 s
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C
3.8 s
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D
4.2 s
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Solution

## The correct option is B 3.4 sLet H be the total height through which the body falls . Time taken by the body to fall through height H t=√2Hg ...(i) It is given that the body falls H/2 distance in time (t−1). Thus we have, t−1=√2(H2)g..(ii) Using (i) and (ii), we have √2Hg−√Hg=1 ⟹√Hg(√2−1)=1 ⟹√Hg=10.414 Given H=58 m ∴ Time of fall is t=√2Hg=1.4140.414=3.4 s Alternative solution : Taking downward as positive and total time of fall be T. For AB: s=ut+12at2 H2=12g(T−1)2(1) For AC: s=ut+12at2 H=12gT2(2) Dividing equation (2) by (1), we get 2=T2(T−1)2 1.414=TT−1 0.414T=1.414 ⇒T=3.4 s

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