A) 26s
B) 25s
C) 13s
D) 12s
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 \[n\,th\]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, \[{{s}_{n}}=s\] \[\therefore \] \[\frac{1}{2}g(2n-1)=\frac{25}{2}\times g\] \[\Rightarrow \]\[2n-1=25\] \[\Rightarrow \]\[n=13\,s\]
You need to login to perform this action.
You will be redirected in
3 sec
