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.
BETA
