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)

5 s

Answer 2

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 $t=5 \,s$