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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Application in Mechanics and Rate Measurer

  • question_answer
    The equations of motion of two stones thrown vertically upwards simultaneously are \[s=19.6\,t-4.9\,{{t}^{2}}\] and \[s=9.8\,t-4.9\,{{t}^{2}}\] respectively and the maximum height attained by the first one is h. When the height of the first stone is maximum, the height of the second stone will be

    A)            h/3

    B)            2h

    C)            h    

    D)            0

    Correct Answer: D

    Solution :

               The time taken by first stone to secure maximum height =\[t=\frac{u}{g}=\frac{19.6}{9.8}=2sec\].                    The time taken by second stone to secure maximum height is, \[t=\frac{9.8}{9.8}=\]1sec. Therefore, in 2 sec, second stone will come back to the ground. Hence height \[s=a{{t}^{2}}+bt+6\].


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