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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    A ball is dropped from a height of 48 m and rebounds\[\frac{2}{3}\]of the distance it falls. If it continues to fall and rebound in this way, the distance that the ball travels before coming to rest is:

    A)  144m                                   

    B)  240m

    C)  120m                   

    D)         96m

    E)  320m

    Correct Answer: B

    Solution :

    \[\because \]\[\frac{2}{3}\times 48={{e}^{2}}=48\] \[\Rightarrow \]               \[e=\sqrt{\frac{2}{3}}\] \[\therefore \]Total travelled distance                 \[=\frac{1+\frac{2}{3}}{1-\frac{2}{3}}\times 48\]                 \[=\frac{\frac{5}{3}}{\frac{1}{3}}\times 48=5\times 48\] \[=240\text{ }m\]


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