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Uttarakhand PMT Uttarakhand PMT Solved Paper-2005

  • question_answer
    A bomb is fired from a cannon with a velocity of of 1000 m/s making an angle of\[30{}^\circ \] with the horizontal. The time taken by the bomb to reach the highest point, will be:

    A)  51 sec            

    B)  25.5 sec

    C)  61 sec             

    D)  21 sec

    Correct Answer: A

    Solution :

     Here: Velocity of bomb \[\upsilon =1000\text{ }m/s\] The angle      \[\theta =30{}^\circ \] Time taken by the bomb to reach the highest point \[t=\frac{\upsilon \sin \theta }{g}\] \[=\frac{1000\sin 30{}^\circ }{9.8}\] So, \[t=\frac{1000\times 0.5}{9.8}=51\,\sec \]


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