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Haryana PMT Haryana PMT Solved Paper-2003

  • question_answer
    As stone falls freely such that the distance           covered by it in the last second of its motion is   equal to the distance covered by it in the first                5 seconds. It is in air for :

    A)  26s                                       

    B)  25s

    C)  13s                                       

    D)  12s

    Correct Answer: C

    Solution :

                    Distance travelled in nth sec \[{{s}_{n}}=u+\frac{1}{2}g(2n-1)\] In case of freely falling body u = 0 So,          \[{{s}_{n}}=\frac{1}{2}g(2n-1)\]           ...(1) Now distance travelled in t sec, is                 \[s=\frac{1}{2}g\times {{(5)}^{2}}=\frac{25}{2}g\]      ??(2) When\[t=5\text{ }sec\], then \[s=\frac{1}{2}g\times {{(5)}^{2}}=\frac{25}{2}g\]       ...(2) According to the condition \[{{s}_{n}}=s\], so, .from eqs. (1) and (2)                 \[\frac{1}{2}g(2n-1)=\frac{g}{2}\times 25\] or            \[2n-1=25\] or            \[n=13\sec \]


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