CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    A ball which is at rest is dropped from a height h metre. As it bounces off the floor its speed is 80% of what it was just before touching the ground. The ball will then rise to nearly a height:

    A)  0.94 h                                  

    B)  0.80 h

    C)  0.75 h                                  

    D)  0.64 h

    E)  0.50 h

    Correct Answer: D

    Solution :

                  \[v=\sqrt{2gh}\]                            ...(1) After rebounce, \[{{v}^{2}}={{u}^{2}}-2gh\] \[\Rightarrow \]               \[{{u}^{2}}={{v}^{2}}+2gh\] \[\therefore \]  \[{{u}^{2}}=2gh\]                                           ...(2) \[\therefore \]  \[\frac{{{v}^{2}}}{{{u}^{2}}}=\frac{2gh}{2gh}\] \[\Rightarrow \]               \[h=h\times \frac{{{u}^{2}}}{{{v}^{2}}}=h\times {{\left( \frac{80}{100} \right)}^{2}}\]                                 \[=0.64\,h\]       


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