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 is

• A. 6 s
• B. 5 s
• C. 4 s
• D. 3 s

Answer 1

Answer: (B)

The distance covered in $3s$,
$s_3$ $=ut$ $+\frac{1}{2}g{t}^2$
or $s_3=0\times t+\frac{1}{2}9.8\times$ $3^2$
or $s_3=44.1m$
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 $t=5s$