JEE Main & Advanced Physics Question Bank Oblique Projectile Motion

  • question_answer
    A body of mass m is thrown upwards at an angle q with the horizontal with velocity v.  While rising up the velocity of the mass after t seconds will be                                [AMU (Engg.) 1999]

    A)  \[\sqrt{{{(v\,\cos \,\theta )}^{2}}+{{(v\,\sin \,\theta )}^{2}}}\]  

    B)             \[\sqrt{{{(v\,\cos \,\theta -v\sin \,\theta )}^{2}}-\,gt}\]

    C)             \[\sqrt{{{v}^{2}}+{{g}^{2}}{{t}^{2}}-(2\,v\,\sin \,\theta )\,gt}\]        

    D)             \[\sqrt{{{v}^{2}}+{{g}^{2}}{{t}^{2}}-(2\,v\,\cos \,\theta )\,gt}\]

    Correct Answer: C

    Solution :

                    Instantaneous velocity of rising mass after t sec will be \[{{v}_{t}}=\sqrt{v_{x}^{2}+v_{y}^{2}}\] where \[{{v}_{x}}=v\cos \theta =\]Horizontal component of velocity \[{{v}_{y}}=v\sin \theta -gt=\]Vertical component of velocity \[{{v}_{t}}=\sqrt{{{(v\cos \theta )}^{2}}+{{(v\sin \theta -gt)}^{2}}}\] \[{{v}_{t}}=\sqrt{{{v}^{2}}+{{g}^{2}}{{t}^{2}}-2v\sin \theta \,gt}\]         


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