A) \[26\,\,s\]
B) \[25\,\,s\]
C) \[13\,\,s\]
D) \[12\,\,s\]
Correct Answer: C
Solution :
Key Idea: In case .of freely falling body initial velocity is zero. From equation of motion the distance travelled in the nth second of motion is \[{{s}_{n}}=u+\frac{1}{2}g(2n-1)\] ... (i) Also\[,\] \[s=ut+\frac{1}{2}g{{t}^{2}}\] ... (ii) Putting \[u=0\] in Eqs. (i) and (ii) and \[t=5\,\,s\] in Eq. (ii), we get \[s=\frac{1}{2}g{{(5)}^{2}}=\frac{25}{2}g,\,\,{{s}_{n}}=\frac{1}{2}g(2n-1)\] Given,\[{{s}_{n}}=s\] \[\therefore \] \[\frac{1}{2}g(2n-1)=\frac{25}{2}\times g\] \[\Rightarrow \] \[2n-1=25\] \[\Rightarrow \] \[n=13\,\,s\]
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