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JEE Main & Advanced Sample Paper JEE Main Sample Paper-32

  • question_answer
    A particle is projected at an angle \[{{\tan }^{-1}}\left( \frac{1}{\sqrt{3}} \right)\]At a height, velocity of the particle is \[9\hat{i}+3\hat{j}\] (m/s). Find the height \[\left( g=10m/{{s}^{2}} \right)\]

    A) 1.8 cm          

    B)                    0.9 cm

    C) 0.7m   

    D)                    0.11m

    Correct Answer: B

    Solution :

    \[\tan \theta \,=\frac{1}{\sqrt{3}}=\frac{{{u}_{y}}}{{{u}_{x}}}=\frac{{{u}_{y}}}{9}\] (Horizontal velocity will remain same)                 \[{{u}_{y}}=\frac{9}{\sqrt{3}}=3\sqrt{3}(m/\sec )\] \[u_{y}^{2}=v_{y}^{2}+2gh\] \[27={{(3)}^{2}}+2\times 10\times h(\sin ce\,{{v}_{y}}=3m/s)\] \[h=\frac{18}{20}=0.90cm\]


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