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JEE Main & Advanced Sample Paper JEE Main Sample Paper-21

  • question_answer
    A particle moves in a projectile motion. At a point Q when its velocity vector makes an angle of \[{{30}^{o}}\]with horizontal, radius of curvature of path at Q is \[\frac{80}{3\sqrt{3}}\](in metres). If the particle is projected at \[{{60}^{o}},\] find the velocity of projection (m/sec)

    A)  \[20\sqrt{3}\]                           

    B)  \[\frac{20}{\sqrt{3}}\]

    C)  20                               

    D)  \[20\sqrt{2}\]

    Correct Answer: C

    Solution :

    component of acceleration g of the projectile perpendicular to the velocity is radial acceleration Hence at Q \[g\cos {{30}^{0}}\,=\frac{{{v}^{2}}}{R}\] \[10\times \frac{\sqrt{3}}{2}\,=\frac{{{v}^{2}}3\sqrt{3}}{80}\] \[v=\frac{20}{\sqrt{3}}\,=(m/\sec )\] Now, \[u\cos {{60}^{0}}\,=v\,\cos {{30}^{0}}\] \[\frac{u}{2}\,=\frac{20}{\sqrt{3}}.\frac{\sqrt{3}}{2}\] \[u=20\,m/\sec \]


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