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.
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