question_answer

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 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 :

[c] 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

\[=h-\frac{h}{9}=\frac{8h}{9}m\]