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JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Conservation of Energy and Momentum

  • question_answer
    A body of mass 2 kg is thrown up vertically with K.E. of 490 joules. If the acceleration due to gravity is 9.8\[m/{{s}^{2}}\], then the height at which the K.E. of the body becomes half its original value is given by                 [EAMCET 1986]

    A)                 50 m     

    B)                 12.5 m

    C)                 25 m     

    D)                 10 m

    Correct Answer: B

    Solution :

                        Let h is that height at which the kinetic energy of the body becomes half its original value i.e. half of its kinetic energy will convert into potential energy                 \ mgh = \[\frac{490}{2}\]Þ \[2\times 9.8\times h=\frac{490}{2}\]Þ \[h=12.5m.\]


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