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Manipal Engineering Manipal Engineering Solved Paper-2012

  • question_answer
    A balloon is rising vertically up with a velocity of\[29\,\,m{{s}^{-1}}\]. A stone is dropped from it and it reaches the ground in\[10\,\,s\]. The height of the balloon when the stone was dropped from it is \[(g=9.8m{{s}^{-1}})\]

    A)  400 m                                  

    B)  150 m

    C)  100 m                  

    D)         200 m

    Correct Answer: D

    Solution :

    For a stone which is thrown downwards from a balloon rising upwards, the equation of motion is                 \[h=-ut+\frac{1}{2}g{{t}^{2}}\]                   \[=-29\times 10+\frac{1}{2}\times 9.8\times {{(10)}^{2}}\]                   \[=-290+490\]                      \[=200\,\,m\]


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