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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2006

  • question_answer
        A body dropped from top of a tower fall through 60 m during the last two seconds of its fall. The height of tower is\[(g=10\text{ }m/{{s}^{2}})\]

    A)  95 m                                    

    B)  60 m

    C)  80 m                                    

    D)  90 m

    Correct Answer: C

    Solution :

                    Let body reaches the ground in t seconds. \[\therefore \]Velocity of body after\[(t-2)\]seconds from equation of motion                 \[v=u+gt\] where v is final velocity, u the initial velocity (= 0) and \[t=t-2\] \[\therefore \]  \[v=g(t-2)\] Distance covered in last two seconds \[h=g(t-2)\times 2+\frac{1}{2}g{{(2)}^{2}}\] \[60=20(t-2)+20\] On solving \[t=4\text{ }s\] Hence, height of tower is given by                 \[h=ut+\frac{1}{2}g{{t}^{2}}\] Since,    \[u=0\] \[\therefore \]  \[h=\frac{1}{2}g{{t}^{2}}\]                 \[=\frac{1}{2}\times 10\times {{(4)}^{2}}=80\,m\].


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