CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the acceleration due to gravity on the surface of the earth. If\[{{R}_{e}}\]is the maximum range of a projectile on the earths surface, what is the maximum range on the surface of the moon for the same velocity of projection?

    A)  \[0.2{{R}_{e}}\]               

    B)         \[2{{R}_{e}}\]                  

    C)  \[0.5{{R}_{e}}\]                               

    D)  Zero

    E)  \[5{{R}_{e}}\]

    Correct Answer: E

    Solution :

    Maximum range \[{{R}_{\max }}=\frac{{{u}^{2}}}{g}\Rightarrow {{R}_{\max }}\propto \frac{1}{g}\] \[\therefore \]  \[\frac{R{{}_{\max }}}{{{R}_{\max }}}=\frac{{{g}_{e}}}{{{g}_{m}}}=\frac{{{g}_{e}}}{0.2\,{{g}_{e}}}\] \[\Rightarrow \]               \[R_{\max }^{}=5{{R}_{e}}\]


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