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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2011

  • question_answer
    A body is projected with kinetic energy T at the angle of maximum range. Its kinetic energy at the highest point of the trajectory will be

    A)  T                                            

    B)  \[\text{T/}\sqrt{\text{2}}\]

    C)  T/2                                       

    D)  zero

    Correct Answer: C

    Solution :

    We know that                 \[T=\frac{1}{2}m{{u}^{2}}\] then,     \[KE=\frac{1}{2}m{{\left( \frac{u}{\sqrt{2}} \right)}^{2}}\]                 \[\theta ={{45}^{o}}\]                    \[KE=\frac{1}{2}T\]


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