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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2011

  • question_answer
    The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by \[y=8t-5{{t}^{2}}m\] and \[x\text{ }=\text{ }6t\]m, where t is in seconds. The velocity with which the projectile is projected is

    A) \[6\text{ }m{{s}^{-1}}\]

    B)  \[8\text{ }m{{s}^{-1}}\]

    C)  \[10\text{ }m{{s}^{-1}}\]

    D)  \[14\text{ }m{{s}^{-1}}\]

    Correct Answer: C

    Solution :

    Given,              \[y=8t-5{{t}^{2}}\]         ...(i) \[x=6t\] ...(ii) We know, \[x=(u\,cos\theta )t\] ...(iii) Compare with Eq. (ii), we get \[{{u}_{1}}\,cos\theta =\frac{x}{t}=6\] and         \[y=(u\sin \theta )t-\frac{1}{2}g{{t}^{2}}\] compare with Eq (i), we get \[{{u}_{2}}\sin \theta =8\] \[\therefore \] \[u=\sqrt{u_{1}^{2}+u_{2}^{2}}\] \[u=\sqrt{36+64}\] \[u=10\,m{{s}^{-1}}\]


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