2. Solved Papers
  3. BVP Medical

BVP Medical BVP Medical Solved Paper-2013

  • question_answer
    A particle is projected with a velocity v such that its range on the horizontal plane is twice the greater height attained by it. The range of the projectile is (where g is acceleration due to gravity)

    A)  \[\frac{4{{v}^{2}}}{\sqrt{5}g}\]                                 

    B)  \[(g=10m/{{s}^{2}})\]

    C)  \[800N\]                                            

    D)  \[1200N\]

    Correct Answer: A

    Solution :

                    \[-\frac{\pi }{3}\] \[\frac{\pi }{6}\] \[-\frac{\pi }{6}\] From triangle we can say that                 \[\frac{\pi }{3}\]                 \[1:2:3\] Range of projectile                 \[1:4:9\]                 \[1:3:5\]


You need to login to perform this action.
You will be redirected in 3 sec spinner