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 is........$sec$
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 is........$sec$
$6$
$5$
$4$
$3$
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 is........$sec$
The distance traveled in last second.
${S_{{\rm{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,
${S_{{\rm{Three}}}} = 0 + \frac{1}{2} \times 9.8 \times 9 = 44.1\;m$
According to problem ${S_{{\rm{Last}}}} = {S_{{\rm{Three}}}}$
$ \Rightarrow 4.9(2t - 1) = 44.1 \Rightarrow 2t - 1 = 9$ $⇒$ $t = 5 sec$.
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