A) \[h/9\]m from the ground
B) \[7h/9\]m from the ground
C) \[8h/9\]m from the ground
D) \[17h/9\]m from the ground
Correct Answer: C
Solution :
Second law of motion gives \[s=ut+\frac{1}{2}g{{T}^{2}}\] Or \[h=0+\frac{1}{2}g{{T}^{2}}\] \[(\because u=0)\] \[\therefore \] \[T=\sqrt{\left( \frac{2h}{g} \right)}\] At \[t=\frac{T}{3}s,\] \[s=0+\frac{1}{2}g{{\left( \frac{T}{3} \right)}^{2}}\] \[\Rightarrow \] \[s=\frac{1}{2}g.\frac{{{T}^{2}}}{9}\] \[\Rightarrow \] \[s=\frac{g}{18}\times \frac{2h}{g}\] \[\left( \because T=\sqrt{\frac{2h}{g}} \right)\] \[\therefore \] \[s=\frac{h}{9}m\] Hence, the position of ball from the ground \[=h-\frac{h}{9}=\frac{8h}{9}m\]You need to login to perform this action.
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