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JEE Main & Advanced Sample Paper JEE Main Sample Paper-13

  • question_answer
    A particle is projected from ground making an angle \[\theta \] with the horizontal. The value of \[\theta \] for which, at the highest point of its trajectory, kinetic energy of particle will be equal to its potential energy, is

    A)  \[{{30}^{o}}\]                                   

    B)  \[{{45}^{o}}\]               

    C)  \[{{60}^{o}}\]                                   

    D)  \[{{75}^{o}}\]

    Correct Answer: B

    Solution :

     Let, F be the KE at the instant of projection, then \[=(2mv)\] As KE at topmost point = PE at topmost point \[=(2mv)\,(nt)\] KE at topmost point, \[({{d}_{d}}g)t=2mv\times nt\] But          \[\Rightarrow \] \[10\times 9.8=2\times 5\times v\times 10\]      \[v=\frac{98}{100}=0.98\,m/s\] \[x={{(t-3)}^{2}}={{t}^{2}}-6t+9\]              \[\Rightarrow \]


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