2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced AIEEE Solved Paper-2006

  • question_answer
    A body falling .from rest under gravity passes a certain point P. It was at a distance of 400 m from P, 4 s prior to passing through P. If \[g=10m/{{s}^{2}},\]then the height above the point P from where the body began to fall is                    AIEEE  Solved  Paper-2006

    A) 900m

    B)                                        320 m   

    C)        680 m                   

    D)        720 m      

    Correct Answer: D

    Solution :

        Let the body be at a height\[{{h}_{1}}\]at a time\[t\]and is at height h at a time\[(t-4)\]from above. \[\therefore \] \[{{h}_{1}}-h=400\] \[\Rightarrow \] \[\frac{1}{2}g{{t}^{2}}-\frac{1}{2}g{{(t-4)}^{2}}=400\] \[\Rightarrow \] \[{{t}^{2}}-{{(t-4)}^{2}}=80\] \[\Rightarrow \] \[(2t-4)4=80\] \[\Rightarrow \] \[t=12\,s\] \[\therefore \] \[h=\frac{1}{2}g{{(t-4)}^{2}}=320\,m\] Hence, total distance of point P from the point from where body began to fall                   \[=320+400=720m\]       


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