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 sec
• B. 5 sec
• C. 4 sec
• D. 3 sec

Answer 1

Answer: (B)

The distance covered in $3\, \sec$,
$s_{3}=u t+\frac{1}{2} g t^{2}$
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
$s_{1} =u+\frac{1}{2} g(2 t-1)$
$44.1 =\theta+\frac{1}{2} \times 9.8(2 t-1)$
$=4.9(2 t-1)$
As both the distances are equal then
$4.9(2 t-1) =44.1$
or $2 t-1 =\frac{44.1}{4.9}=9$
or $2 t =9+1 \cdot=10$
or $t =5\, \sec$