MGIMS WARDHA MGIMS WARDHA Solved Paper-2014

  • question_answer
    A ball is dropped from a tower of height h. The duration\[({{t}_{0}})\]of motion, when it reaches bottom of the tower is given by

    A)  \[\sqrt{\frac{2h}{g}}\]                  

    B)  \[\sqrt{\frac{h}{g}}\]

    C)  \[2\sqrt{\frac{h}{g}}\]                  

    D)  \[4\sqrt{\frac{h}{g}}\]

    Correct Answer: A

    Solution :

                    From the equation of motion \[h=ut+\frac{1}{2}g{{t}^{2}},\] We get \[h=0+\frac{1}{2}gt_{0}^{2}\]                 \[{{t}_{0}}=\sqrt{2h/g}\]


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