Question

The ratio of traveled distances by freely falling body in first, second and third seconds will be

• A. 5 : 3 : 1
• B. 1 : 4 : 9
• C. 1 : 3 : 5
• D. 9 : 4 : 1

Answer 1

Answer: (C)

Travelling distance in first second,
$<br/>s_1=0\times 1+\frac{1}{2}g\times (1{)}^2=\frac{1}{2}g<br/>$
After one second achieve velocity,
$<br/>v=0+g\times 1=g<br/>$
Travelling distance after $2^{\text{th }}$ second,
$<br/>s_2=g\times 1+\frac{1}{2}g\times (1{)}^2=\frac{3}{2}g<br/>$
Velocity, $V=g+g(1{)}^2=2g$
Travelling distance in third second,
$<br/>s_3=2g\times 1+\frac{1}{2}g(1{)}^2=\frac{5}{2}g<br/>$
Thus $s_1:s_2:s_3=1:3:5$