A) 12 sec
B) 13 sec
C) 25 sec
D) 26 sec
Correct Answer: B
Solution :
Distance travelled in nth second \[{{s}_{n}}=u+\frac{1}{2}g(2n-1)\] For freely falling body \[u=0\] \[\therefore \] \[{{s}_{n}}=\frac{1}{2}g(2n-1)\] Distance travelled in t sec. \[s=ut+\frac{1}{2}g{{t}^{2}},\]put \[t=5\] \[\therefore \] \[s=\frac{1}{2}g{{t}^{2}}=\frac{g\times 25}{2}\] Now, according to the question \[{{s}_{n}}=s\] \[\therefore \] \[\frac{1}{2}g(2n-1)=\frac{g}{2}\times 25\] So, \[n=13\,\sec \]You need to login to perform this action.
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