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  4. 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

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Q. 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

Punjab PMETPunjab PMET 2005Motion in a Straight Line

Solution:

The distance covered in $3 \,s$,
$s_3$ $=ut$ $+\frac{1}{2}\,gt^2$
or $s_3=0 \times t +\frac{1}{2}9.8 \times$ $3^2$
or $s_3 =44.1 m$
The distance covered in last second $s_1=u+\frac{1}{2}g(2t-1)$
$44.1=0+\frac{1}{2}\times 9.8 (2t-1)$ =4.9(2t-1)
As both the distances are equal then
$4.9 (2t -1)=44.1$
$2t-1=\frac{44.1}{4.9}=9$
or $2t =9+ 1 = 10$ or $t=5 \,s$