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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    A ball is projected from the ground at a speed of\[10\text{ }m{{s}^{-1}}\]making an angle of\[30{}^\circ \]with the horizontal. Another ball is simultaneously released from a point on the vertical line along the maximum height of the projectile. The initial height of the second ball is \[(g=10m{{s}^{-2}})\]

    A)  6.25m                                  

    B)  2.5m    

    C)         3.75m                  

    D)         5m

    E)  1.25m

    Correct Answer: B

    Solution :

    Maximum height of projectile \[h=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[\therefore \]  \[h=\frac{{{(10)}^{2}}\times {{\sin }^{2}}(30{}^\circ )}{2\times 10}\]                 \[=\frac{5}{4}=1.25\,m\] Time to reach maximum height                 \[t=\frac{u\sin \theta }{g}\] \[\therefore \]  \[t=\frac{10\times \sin 30{}^\circ }{10}=0.5\,s\] So, distance of vertical fall in 0.5 s                 \[s=\frac{1}{2}g{{t}^{2}}\] or    \[s=\frac{1}{2}\times 10\times {{(0.5)}^{2}}=1.25\,m\] \[\therefore \]  Height of second ball \[=1.25+1.25=2.5m\]


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