2. Solved Papers
  3. CET Karnataka Engineering

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


You need to login to perform this action.
You will be redirected in 3 sec spinner