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BCECE Engineering BCECE Engineering Solved Paper-2006

  • question_answer
    A particle is dropped under gravity from rest from a height \[h(g=9.8\,m/{{s}^{2}})\]and it travels a distance \[\frac{9h}{25}\] in the last second the height \[h\]is:

    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,    \[h=\frac{1}{2}\times 9.8\times 5\times 5\] \[\Rightarrow \]               \[h=122.5\,m/s\]


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