question_answer

    A ball is released from the top of a tower of height h m. It takes T s to reach the ground. What is the position of the ball in 773 s?

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