A) 15 m
B) 20 m
C) 25 m
D) 30 m
Correct Answer: C
Solution :
\[s=ut+\frac{1}{2}a{{t}^{2}}\] For 1st body \[u=0\]and\[a=g\][freely falling body] Distance covered in 2 s, \[{{s}_{1}}=0+\frac{1}{2}g{{(2)}^{2}}\] For 2nd body Distance covered in 2 s, \[{{s}_{2}}=0+\frac{1}{2}g{{(2)}^{2}}\] \[\therefore \] \[{{s}_{2}}-{{s}_{1}}=\frac{1}{2}g[{{(3)}^{2}}-{{(2)}^{2}}]\] \[=\frac{1}{2}g[9-4]\] \[=25\,m\]
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