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Manipal Medical Manipal Medical Solved Paper-2000

  • question_answer
    If a freely falling body travels in the last second a distance equal to the distance travelled by it in the first three second, the time of the travel is:

    A)  6 sec              

    B)  5 sec

    C)  4 sec              

    D)  3 sec

    Correct Answer: B

    Solution :

     The distance covered in 3 sec, \[{{s}_{3}}=ut+\frac{1}{2}g{{t}^{2}}\] Or \[{{s}_{3}}=0\times t+\frac{1}{2}9.8\times {{3}^{2}}\] Or \[{{s}_{3}}=44.1\,m\] The distance covered in last second \[{{s}_{1}}=u+\frac{1}{2}g(2t-1)\] \[44.1=\theta +\frac{1}{2}\times 9.8(2t-1)\] \[=4.9(2t-1)\] As both the distances are equal then \[4.9(2t-1)=44.1\] \[2t-1=\frac{44.1}{4.9}=9\] Or \[2t=9+1=10\] Or       \[t=5\text{ }sec\]


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