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JCECE Engineering JCECE Engineering Solved Paper-2014

  • question_answer
    A stone is thrown vertically upwards. When the stone is at a height equal to half of its maximum height its speed will be\[10\,\,m/s\], then the maximum height attained by the stone (Take\[g=10\,\,m/{{s}^{2}})\]

    A) \[3\,\,m\]                                           

    B) \[15\,\,m\]

    C) \[1\,\,m\]                                           

    D) \[10\,\,m\]

    Correct Answer: D

    Solution :

    Let \[u\] be the initial velocity and \[h\] be the maximum height attained by the stone                 \[v_{1}^{2}={{u}^{2}}-2gh\]                 \[{{(10)}^{2}}={{u}^{2}}-2\times 10\times \frac{h}{2}\]                   ... (i) Again at height\[h\]                 \[v_{2}^{2}={{u}^{2}}-2gh\]                 \[{{(0)}^{2}}={{u}^{2}}-2\times 10\times h\]                 \[{{u}^{2}}=20h\]                                             ... (ii) So, from Eqs. (i) and (ii), we have                 \[100=10h\]                 \[h=10m\]


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