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

  • question_answer
    The maximum height is reached in 5 seconds by a stone thrown vertically upwards and moving under the equation 10s = 10ut ? 49\[{{t}^{2}}\], where s is in metre and t is in second. The value of u is

    A)            \[4.9m/\sec \]

    B)            \[49m/\sec \]

    C)            \[98m/\sec \]

    D)            None of these

    Correct Answer: B

    Solution :

               Given equation is\[\frac{dy}{dt}=-\frac{16}{6}=-\frac{8}{3}cm/\sec .\] or \[s=ut-4.9{{t}^{2}}\]            Þ \[\frac{ds}{dt}=u-9.8t=v\]                    When stone reached the maximum height, then \[v=0\]                    Þ \[u-9.8t=0\Rightarrow u=9.8t\]                    But time \[t=5\]sec                    So the value of \[u=9.8\times 5=49.0\]m/sec                    Hence initial velocity \[=49\]m/sec.


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