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
A) $6\ sec$
B) $5\ sec$
C) $4\ sec$
D) $3\ sec$
Explanation
(b) 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 $S_{Last} = S_{Three}$
$\Rightarrow 4.9(2t - 1) = 44.1 \Rightarrow 2t - 1 = 9$
$\Rightarrow t = 5\ sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.