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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2011

  • question_answer
    A body of mass 5 kg is thrown vertically up with a kinetic energy of 490 J. The height at which the kinetic energy of the body becomes half of the original value is (acceleration due to gravity\[=9.8\text{ }m{{s}^{-2}}\])

    A)  5 m            

    B)  2.5 m

    C)  10 m           

    D)  12.5 m

    Correct Answer: A

    Solution :

    Given,     \[m=5\,kg\]   and             KE = 490 J By the law of conservation of energy \[\frac{1}{2}m{{u}^{2}}=\frac{1}{2}m{{v}^{2}}+mgh\] \[490=245+5\times 9.8\times h\] \[h=\frac{490-245}{5\times 9.8}\] \[h=\frac{245}{49}\] \[=5m\]


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