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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

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

    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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