A) 100 m
B) 122.5 m
C) 145 m
D) 167.5 m
Correct Answer: B
Solution :
Given, particle is dropped under gravity from rest of height h and \[g=9.8\,m/{{s}^{2}}\]and travels a distance \[\frac{9h}{25}\] in the last second. \[\therefore \] \[{{h}_{nth}}=u+\frac{1}{2}g(2n-1)\] \[=\frac{1}{2}g(2n-1)\] \[(\because \,u=0)\] \[\Rightarrow \] \[\frac{9h}{25}=\frac{1}{2}g(2n-1)\] But \[h=\frac{1}{2}g\,{{n}^{2}}\] \[\Rightarrow \] \[\frac{9}{25}\times \frac{1}{2}g{{n}^{2}}=\frac{1}{2}g(2n-1)\] \[\Rightarrow \] \[9{{n}^{2}}-50n+25=0\] \[\Rightarrow \] \[n=5,\frac{5}{9}\] But \[n=\frac{5}{9}\]is not possible. \[\therefore \] \[n=5\,s\] Then, \[h=\frac{1}{2}\times 9.8\times 5\times 5\] \[\Rightarrow \] \[h=122.5\,m/s\]You need to login to perform this action.
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