BCECE Engineering BCECE Engineering Solved Paper-2002

  • question_answer
    A stone is thrown with an initial speed of 4.9 m/s from a bridge in vertically upward direction. It falls down in water after 2 s. The height of the bridge is:

    A)  24.7 m                                 

    B)  19.8 m

    C)  9.8 m                   

    D)         4.9 m

    Correct Answer: C

    Solution :

    From Newtons second equation of motion, we have \[s=ut+\frac{1}{2}g{{t}^{2}}\] where\[u\]is initial velocity,\[t\]is time, s is displacement, and g is the acceleration due to gravity. Since, gravity attract everybody towards its centre, downward direction is taken as positive and upward negative.                 Hence, \[s=ut+\frac{1}{2}g{{t}^{2}}\]                                 \[=(-4.9\times 2)+\frac{1}{2}\times 9.8\times {{2}^{2}}\]                                 \[=-9.8+19.6=9.8\,m\]


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