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NEET NEET SOLVED PAPER 2014

  • question_answer
    A projectile is fired from the surface of the earth with a velocity of 5 \[m{{s}^{-1}}\] and angle \[\theta \]with the horizontal. Another projectile fired from another planet with a velocity of 3 \[m{{s}^{-1}}\] at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in\[m{{s}^{-2}}\]) is (given, \[g=9.8\,m{{s}^{-2}}\]) [AIPMT 2014]

    A)  3.5   

    B)  5.9    

    C)  16.3   

    D)  110.8

    Correct Answer: A

    Solution :

    If the trajectory is same for both the projectiles, their maximum height will be same. \[{{({{H}_{\max }})}_{1}}={{({{H}_{\max }})}_{2}}\] \[\frac{u_{1}^{2}{{\sin }^{2}}\theta }{2{{g}_{1}}}=\frac{u_{2}^{2}{{\sin }^{2}}\theta }{2{{g}_{2}}}\] \[\frac{u_{1}^{2}}{u_{2}^{2}}=\frac{{{g}_{1}}}{{{g}_{2}}}\Rightarrow \frac{{{(5)}^{2}}}{{{(3)}^{2}}}=\frac{9.8}{{{g}_{2}}}\] \[{{g}_{2}}=\frac{9.8\times 9}{25}=3.5m/{{s}^{2}}\]


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