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

  • question_answer
     A ball thrown vertically upwards falls back on the ground after 6 second. Assuming that the equation of motion is of the form \[s=ut-4.9{{t}^{2}}\], where s is in metre and t is in second, find the velocity at \[t=0\]

    A)            \[0\,m/s\]

    B)            1 m/s

    C)            29.4 m/s

    D)            None of these

    Correct Answer: C

    Solution :

               Velocity of ball (v) =\[\frac{ds}{dt}=\]\[u-9.8t\]                    Here terminal velocity \[v=0\]and \[t=3\,sec\]                                                               \[u=9.8(3)=29.4m/sec\].


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