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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    A plane is inclined at an angle \[\alpha =30{}^\circ \]with a respect to the horizontal. A particle is projected with a speed \[u=2m{{s}^{-1}}\]from the base of the plane, making an angle \[\theta ={{15}^{o}}\]with respect to the plane as shown in the figure. The distance from the base, at which the particle hits the plane is close to : (Take \[g=10\,m{{s}^{-2}}\]) [JEE Main 10-4-2019 Afternoon]

    A) 14 cm             

    B) 20 cm

    C) 18 cm             

    D) 26 cm

    Correct Answer: B

    Solution :

    \[t=\frac{2\times 2\times \sin {{15}^{o}}}{g\,\cos {{30}^{o}}}\]             \[S=2\cos {{15}^{o}}\times t-\frac{1}{2}g\sin {{30}^{o}}{{t}^{2}}\] Put values and solve \[S\simeq 20cm\]


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