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Q.
The ratio of traveled distances by freely falling body in first, second and third seconds will be
Rajasthan PETRajasthan PET 2001
Solution:
Travelling distance in first second,
$
s_{1}=0 \times 1+\frac{1}{2} g \times(1)^{2}=\frac{1}{2} g
$
After one second achieve velocity,
$
v=0+g \times 1=g
$
Travelling distance after $2^{\text {th }}$ second,
$
s_{2}=g \times 1+\frac{1}{2} g \times(1)^{2}=\frac{3}{2} g
$
Velocity, $V=g+g(1)^{2}=2 g$
Travelling distance in third second,
$
s_{3}=2 g \times 1+\frac{1}{2} g(1)^{2}=\frac{5}{2} g
$
Thus $s_{1}: s_{2}: s_{3}=1: 3: 5$