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A ball is released from the top of a tower of height h metre. It takes T second to reach the ground. What is the position of the ball in \[T/3s\]?     AIEEE  Solved  Paper-2004  

A) \[h/9\]m from the ground

B) \[7h/9\]m from the ground

C) \[8h/9\]m from the ground

D) \[17h/18\]m from the ground

Correct Answer: C

Solution :

Second law of motion gives \[s=ut+\frac{1}{2}g{{t}^{2}}\] \[h=0+\frac{1}{2}g{{T}^{2}}\]                      \[(\because u=0)\] \[\therefore \]\[T=\sqrt{\left( \frac{2h}{g} \right)}\] At\[t=\frac{T}{3}s\] \[s=0+\frac{1}{2}g{{\left( \frac{T}{3} \right)}^{2}}\] \[\Rightarrow \]\[s=\frac{1}{2}g.\frac{{{T}^{2}}}{9}\] \[\Rightarrow \]\[s=\frac{g}{18}\times \frac{2h}{g}\]                                      \[\left( \because T=\sqrt{\frac{2h}{g}} \right)\] \[\therefore \]\[s=\frac{h}{g}m\] Hence, the position of ball from the ground \[=h-\frac{h}{9}=\frac{8h}{9}m\]