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JEE Main & Advanced Physics Question Bank Oblique Projectile Motion

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    An object is projected with a velocity of 20 m/s making an angle of \[{{45}^{o}}\] with horizontal. The equation for the trajectory is\[~h=Ax\text{ }\text{ }B{{x}^{2}}\] where h is height, x is horizontal distance, A and B are constants.  The ratio A : B is  (\[~h=AxB{{x}^{2}}\])                               [EAMCET 2001]

    A)             1 : 5    

    B)             5 : 1

    C)             1 : 40  

    D)             40 : 1

    Correct Answer: D

    Solution :

                    Standard equation of projectile motion                         \[y=x\tan \theta -\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }\]                         Comparing with given equation                         \[A=\tan \theta \] and \[B=\frac{g}{2{{u}^{2}}{{\cos }^{2}}\theta }\]                         So \[\frac{A}{B}=\frac{\tan \theta \times 2{{u}^{2}}{{\cos }^{2}}\theta }{g}=40\]             (As \[\theta =45{}^\circ ,\ u=20\,m/s,\ g=10\,m/{{s}^{2}}\])


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