Kinematics Question 246

Question: If a freely falling body travels in the last second a distance equal to the distance travelled by it in the first three second, the time of the travel it’s [Pb. PMT 2004; MH CET 2003]

Options:

A) 6 sec

B) 5 sec

C) 4 sec

D) 3 sec

Show Answer

Answer:

Correct Answer: B

Solution:

The distance traveled in last second. $ {S _{Last}}=u+\frac{g}{2}(2t-1) $

$ =\frac{1}{2}\times 9.8(2t-1) $

$ =4.9(2t-1) $ and distance traveled in first three second, $ {S _{Three}}=0+\frac{1}{2}\times 9.8\times 9=44.1\ m $ According to problem $ {S _{Last}}={S _{Three}} $
$ \Rightarrow 4.9(2t-1)=44.1\Rightarrow 2t-1=9 $

therefore t = 5 sec.



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