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CET Karnataka Engineering CET - Karnataka Engineering 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 \[y=8\text{ }t5\text{ }{{t}^{2}}m\] and \[x=6t\text{ }m\], 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,                   \[y=8\,t-5\,{{t}^{2}}\]                    ... (i)                                 \[x=6\,t\]                                         .... (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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