CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    From the top of tower, a stone is thrown up. It reaches the ground in\[{{t}_{1}}\]second. A second stone thrown down with the same speed reaches the ground in\[{{t}_{2}}\]second. A third stone released from rest reaches the ground in\[{{t}_{3}}\]second. Then:

    A)  \[{{t}_{3}}=\frac{({{t}_{1}}+{{t}_{2}})}{2}\]                         

    B)  \[{{t}_{3}}=\sqrt{{{t}_{1}}{{t}_{2}}}\]

    C)  \[\frac{1}{{{t}_{3}}}=\frac{1}{{{t}_{1}}}-\frac{1}{{{t}_{2}}}\]        

    D)         \[t_{3}^{2}=t_{2}^{2}-t_{1}^{2}\]

    E)  \[{{t}_{3}}=\frac{({{t}_{1}}-{{t}_{2}})}{2}\]

    Correct Answer: B

    Solution :

    Not available


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